Moving Charges and Magnetism: Class 12 Physics Chapter 4 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

155 min
1 simulation
20 practice questions
40 MCQs

Is chapter mein aap kya seekhenge

Lorentz force aur F = I l × B use karke moving charges aur current-carrying wires par lagne wale magnetic forces ka size aur direction nikaalo
Samjho ki uniform magnetic field mein charge circle ya helix mein kyun chalta hai, aur uska radius, period, cyclotron frequency aur pitch calculate karo
Biot–Savart law state karo aur usse current-carrying ring ke axis par kisi bhi point par field nikaalo
Ampère's circuital law lagakar long straight wire, thick wire aur long solenoid ke fields nikaalo
Do parallel currents ke beech ka force derive karo aur use ampere ki definition se jodo
Current loop par torque derive karo, uska magnetic moment define karo aur samjhao ki chhota loop magnetic dipole ki tarah kyun behave karta hai
Samjhao ki moving coil galvanometer kaise kaam karta hai aur use ammeter ya voltmeter mein kaise badla jaata hai

NCERT Physics Class 12, Chapter 4, pp. 107–135 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Har electric motor, loudspeaker, hospital ka MRI scanner aur particle accelerator ek aise link par depend karta hai jo 1820 tak chhupa raha: electric current magnetic field banata hai. Usse pehle dono subjects ki alag history aur alag toolkits the. Electricity ka matlab tha ragda hua amber, sparks aur Leyden jars; magnetism ka matlab tha lodestones aur woh compass needles jinse sailors raasta dhoondhte the.

Copenhagen ke professor Hans Christian Oersted ko yeh 1820 mein apne students ko ek experiment dikhate hue mila, aur tum ise khud aasaani se dekh sakte ho (Fig. 4.1). Ek wire ko compass ke upar uski needle ke parallel rakho, aur use ek pal ke liye cell se jodo. Needle jhatke se side mein ghoom jaati hai aur jab tak current behta hai, ghoomi hi rehti hai. Connections badal do, to woh doosri taraf ghoomti hai. Current badhane par yeh jhatka badhta hai, aur wire ko needle se upar uthane par kam hota hai.

2/4North ki taraf behta current needle ke north end ko west ki taraf ghuma deta hai.

Fig. 4.1Compass ke theek upar pakdi wire ke saath Oersted ka observation. Needle sirf tab ghoomti hai jab current behta hai, aur uska ghoomna current ki direction aur size par depend karta hai.NCERT Fig. 4.1 · p. 107

Compass ka top view jis par W aur E mark hain, aur ek seedhi copper rang ki wire us par north se south tak, needle ke theek upar, ja rahi hai. Paas mein ek north arrow bana hai. Step 1: current nahi hai aur needle, jiska north end solid aur south end hollow hai, wire ke saath north dikhati hai. Step 2: wire par arrows north ki taraf behta current dikhate hain, aur needle ka north end lagbhag 50 degrees west ki taraf ghoom gaya hai. Step 3: arrows south ki taraf point karte hain, aur needle utna hi east ki taraf ghoom gayi hai. Step 4: double arrows north ki taraf behta dugna current dikhate hain, aur needle aur zyada, lagbhag 70 degrees west ki taraf ghoomti hai.

Needle exactly kahan point karne ki koshish kar rahi hai? Ek current-carrying wire ko card ke beech se seedha khada karo, aur uske chaaron taraf chhote compasses ki ring rakho ya iron filings chhidko. Needles aur filings wire ko centre maan kar circles mein line up ho jaati hain. Yaani straight current ka field apne source se door point nahi karta, jaise charge ka electric field karta hai; yeh uske chaaron taraf lipat jaata hai. To current sirf wire ko garam karne ya bulb jalane se zyada karta hai: yeh apne aas-paas ki jagah ko magnetic field se bhar deta hai. Yahi discovery is chapter ka starting point hai.

Ghar par try karo

Tumhari desk par Oersted

Tumhe chahiye

ek chhota magnetic compass
ek 1.5 V cell (AA ya D size)
lagbhag 40 cm insulated copper wire jiske ends khule hon
sticky tape

Pehle guess karo

Wire needle ke upar hai aur current north ki taraf behta hai. Ab tum wire ko needle ke neeche le jaate ho, current wahi rakhte hue. Needle kya karegi?

Pehle jaisi hi taraf ghoomegiUlti taraf ghoomegiNorth hi dikhati rahegi

Compass ko lakdi ki table par flat rakho, steel aur electrical cables se door, aur needle ko rukne do.
Wire ko compass ke upar seedha, needle ke parallel failao, aur dono taraf tape se chipka do.
Wire ke dono ends ko cell ke terminals se sirf lagbhag ek second ke liye chhuao, aur needle ke north end ko dekho.
Cell par connections badlo aur dohrao.
Ab wire ko compass ke upar ke bajaye neeche khiskao aur dono tests dohrao.

Table 4.1

Wire ki position	Current ki direction	North end kidhar ghooma (east ya west?)
upar	north ki taraf
upar	south ki taraf
neeche	north ki taraf
neeche	south ki taraf

Tumhari entries apne aap save hoti hain.

Timeline

Ghoomti needle se electromagnetism tak

1820
Oersted ki needle
Wire mein current paas ki compass needle ko ghuma deta hai; yeh pehla saboot tha ki electricity aur magnetism jude hain.
1820
Biot aur Savart
Magnetised needles ke saath measurements se current ke chhote piece ke field ka law milta hai.
1820–27
Ampère ke experiments
Parallel currents attract karte hain, antiparallel currents repel karte hain; Ampère currents ke beech forces ki mathematical theory banate hain.
1864
Maxwell's equations
Electricity, magnetism aur light ek hi theory, electromagnetic field ki theory, mein saath aate hain.
1887–95
Radio waves
Germany mein Heinrich Hertz radio waves banate aur detect karte hain; Kolkata mein J. C. Bose millimetre-wavelength waves generate karke unki study karte hain, aur Marconi radio signals bhejna shuru karte hain.
1895
Lorentz force
Lorentz electric aur magnetic fields mein moving charge par force ko usi form mein likhte hain jo aaj use hoti hai.
1931
Cyclotron
Ernest Lawrence magnetic field mein charged particles ke circle mein ghoomne ka use karke unhe high energies tak accelerate karte hain.

Poore chapter mein do sawaal chalte hain. Magnetic field moving charge ko, ya current wali wire ko, kaise push karta hai? Aur current sabse pehle magnetic field banata kaise hai? Inke jawab samjhaate hain ki charged particles field mein circle aur spiral kyun karte hain (cyclotron ke peeche ka idea), paas-paas ki wires ek doosre ko kyun kheenchti hain, aur magnet mein ghoomti ek chhoti coil kaise moving coil galvanometer ban jaati hai jisse currents aur voltages naape jaate hain.

Is chapter ke kai diagrams mein currents aur fields seedhe page ke andar ya bahar point karte hain, aur hum unhe archery se liye do symbols se banate hain (Fig. 4.2). Dot wala circle kisi cheez ko page se bahar, tumhari taraf aate dikhata hai, jaise tumhari taraf seedha udta arrow ka nok. Cross wala circle kisi cheez ko page ke andar, tumse door jaate dikhata hai, jaise door udte arrow ki poonch par lage crossed pankh.

Fig. 4.2Dot aur cross: page ke right angle par currents aur fields ke symbols.NCERT p. 108

Left mein ek circle jiske centre mein dot hai, label: page se bahar, tumhari taraf. Right mein ek circle jismein diagonal cross hai, label: page ke andar, tumse door. Dono ke beech ek arrow side mein bana hai, jiska nok left mein hai (label tip) aur crossed poonch ke pankh right mein (label tail); yeh arrow ke har end ko us symbol se jodta hai jaisa woh saamne se dikhta hai.

Quick check

3 mein se 1

Ek wire compass needle ke theek upar hai aur usmein current hai. Current ulta karne par needle ke swing ka kya hota hai?

Woh usi direction mein aur zyada ghoomti haiWoh utna hi doosri taraf ghoomti haiWoh ghoomna band karke north dikhati hai

Answer dekho

Answer (B) Woh utna hi doosri taraf ghoomti hai

Current ulta hone par wire ke around field bhi ulta ho jaata hai, isliye needle doosri taraf ghoomti hai. Woh phir north tabhi dikhati hai jab current off kiya jaaye.

4.2 Magnetic Force

4.2.1 Sources aur fields

Chapter 1 mein humne electric interaction ko do steps mein toda tha. Ek charge $Q$ pehle apne aas-paas ki jagah ko electric field se bhar deta hai; phir doosra charge $q$ apni position par jo field paata hai, us par respond karta hai. Origin par rakhe point charge ke liye position $𝐫$ par field hai

𝐄 = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}} \hat{𝐫}

(4.1)

jahan

$\hat{𝐫}$: $Q$ se point $𝐫$ ki taraf point karne wala unit vector

aur wahan rakhe charge $q$ par force hai

𝐅 = q 𝐄 = \frac{1}{4 π ε_{0}} \frac{q Q}{r^{2}} \hat{𝐫}

(4.2)

Chapter 1 ne humein do kaam ke ideas diye. Pehla hai superposition: kai sources ka field un fields ka vector sum hota hai jo woh ek-ek karke banate. Doosra yeh ki field sirf hisaab-kitaab nahi hai. Jab mobile-phone tower mein charges hilte hain, to tumhare phone ke charges tabhi respond karte hain jab disturbance speed of light se gap paar kar leta hai, aur apne saath energy aur momentum le jaata hai. Magnetic fields mein bhi yeh dono properties hain. Lekin yahan har current steady hai, isliye har field pal-pal same rehta hai aur koi delay nahi aata.

Magnetism bhi inhi do steps mein kaam karta hai. Moving charges, aur isliye currents, ek magnetic field $𝐁 (𝐫)$ (चुंबकीय क्षेत्र) banate hain, jo space ke har point par size aur direction wala vector hai; phir koi bhi doosra moving charge apni position par mile field par respond karta hai. Naya yeh hai ki woh kaise respond karta hai, aur hum wahin se shuru karte hain.

4.2.2 Magnetic field, Lorentz force

Ek charged particle ko aise region se guzaaro jismein electric field $𝐄$ aur magnetic field $𝐁$ dono hon. Electric force jaana-pehchana $q 𝐄$ hai. Magnetic force particle ki velocity $𝐯$ par bhi depend karta hai, cross product ke through. Position $𝐫$ par point charge $q$ par poora force hai

Lorentz force

𝐅 = q [𝐄 (𝐫) + 𝐯 \times 𝐁 (𝐫)] = 𝐅_{electric} + 𝐅_{magnetic}

(4.3)

Definition

Lorentz force

Woh total force $𝐅 = q (𝐄 + 𝐯 \times 𝐁)$ jo electric aur magnetic fields milkar velocity $𝐯$ se chalte point charge $q$ par lagaate hain.

Example $0.5 T$ field ke right angle par $10^{6} m / s$ se chalte proton par $8 \times 10^{- 14} N$ ka magnetic force lagta hai.

Dutch physicist Hendrik Lorentz ke naam wala yeh expression Ampère aur unke baad ke scientists ke kai experiments ka saar hai. Iska magnetic part, $q 𝐯 \times 𝐁$ , electric part se bilkul alag hai:

Ruke hue charge par koi magnetic force nahi lagta, field chahe kitna bhi strong ho. Force sirf tab tak hai jab tak charge chal raha hai.
Force hamesha velocity aur field dono ke right angle par hota hai, positive charge ke liye $𝐯 \times 𝐁$ ki direction mein (Fig. 4.3).
Iska size, $q v B \sin θ$ , $𝐯$ aur $𝐁$ ke beech ke angle $θ$ par depend karta hai. Field lines ke saath, aage ya peeche, chalne par koi force nahi lagta.
Charge ka sign badalne se force ulta ho jaata hai: saath-saath chalte electron aur proton ulti directions mein dhakele jaate hain.

1/4Jab $𝐯$ , $𝐁$ ke right angle par ho to force sabse bada, $q v B$ , hota hai. Yeh $𝐯$ aur $𝐁$ ke plane ke perpendicular hai: $𝐯$ (picture ke andar) se $𝐁$ (right ki taraf) ki taraf ghumane par right-hand thumb neeche jaata hai.

Fig. 4.3Magnetic force

q 𝐯 \times 𝐁

𝐯

aur

𝐁

ke plane ke right angle par hai, aur iska size

q v B \sin θ

hai.NCERT Fig. 4.2 · p. 110

Left mein perspective mein ek flat shaded plane bana hai jis par ek charge hai, plane ke saath magnetic field B ke liye green arrow aur velocity v ke liye blue arrow. Ek orange arrow plane ke right angle par force F dikhata hai; B right ki taraf point karta hai aur v plane ke andar peeche ki taraf jaata dikhaya gaya hai. Right mein ek chhota graph F ko v aur B ke beech ke angle ke against dikhata hai, 0 se 180 degrees tak ek sine curve, jis par ek dot abhi ka angle mark karta hai. Step 1: v, B se 90 degrees par hai, F seedha neeche apni sabse badi length mein point karta hai aur dot curve ke top par hai. Step 2: v, B se 30 degrees par hai, F aadha lamba hai aur dot curve par aadhi oonchai par hai. Step 3: v, B ke saath point karta hai, koi force nahi hai, label F equals 0, aur dot zero par hai. Step 4: charge negative hai, v phir 90 degrees par hai, aur force utne hi size ka hai lekin seedha upar point karta hai.

$𝐯 \times 𝐁$ ki direction nikaalne ke liye apne right hand ki ungliyan $𝐯$ ke saath point karo aur unhe dono ke beech ke chhote angle se $𝐁$ ki taraf mod do. Tab tumhara thumb $𝐯 \times 𝐁$ ki direction dikhata hai. Positive charge ke liye yahi magnetic force ki direction hai; negative charge ke liye force ulti taraf hota hai. Cartesian axes mein cyclic products $\hat{𝐢} \times \hat{𝐣} = \hat{𝐤}$ , $\hat{𝐣} \times \hat{𝐤} = \hat{𝐢}$ aur $\hat{𝐤} \times \hat{𝐢} = \hat{𝐣}$ yahi kaam karte hain.

Yeh formula yeh bhi batata hai ki $B$ kaise naapein. Rearrange karne par $B = F / (q v \sin θ)$ , isliye magnetic field ki SI unit newton per coulomb per (metre per second) hai. Is unit ko tesla (T) kehte hain, engineer Nikola Tesla ke naam par: $1 T$ ka field, use right angle par $1 m / s$ se cross karte $1 C$ charge par $1 N$ ka force lagata hai. Base units mein,

1 T = 1 N s C^{- 1} m^{- 1} = 1 N A^{- 1} m^{- 1}

In units mein roz ke fields bahut chhote hain. Earth ka field, jo compass ghumata hai, sirf kuch guna $10^{- 5} T$ hai, isliye weak fields ke liye log aksar purani, non-SI unit gauss ( $1 G = 10^{- 4} T$ ) use karte hain. Table 4.2 dikhata hai ki magnetic fields ki range kitni badi hai.

Table 4.2Kuch magnetic field strengths (orders of magnitude)

Source	Field B (tesla)
Neutron star ki surface	lagbhag $10^{8}$
Sabse strong steady laboratory magnets	lagbhag $45$
Hospital ka MRI scanner	$1.5$ se $3$
Chhote fridge magnet ke bilkul paas	lagbhag $10^{- 2}$
Surface par Earth ka field	lagbhag $5 \times 10^{- 5}$
Insaan ke brain se aane wale magnetic signals	lagbhag $10^{- 12}$

Quick check

4 mein se 1

Ek proton (

q = 1.6 \times 10^{- 19} C

)

0.20 T

ke uniform field ke right angle par

5.0 \times 10^{5} m / s

se chalta hai. Us par magnetic force ka size kitna hai, newtons mein?

Answer dekho

Answer 1.6e-14 N

F = q v B \sin 90^{\circ} = (1.6 \times 10^{- 19}) (5.0 \times 10^{5}) (0.20) = 1.6 \times 10^{- 14} N

4.2.3 Current-carrying conductor par magnetic force

Current wali wire moving charges se bhari hoti hai, isliye magnetic field poori wire ko push karega hi. Ek seedhi rod lo, length $l$ aur cross-section $A$ , jo uniform external field $𝐁$ mein padi hai (Fig. 4.4). Maan lo, metal ki tarah, ismein ek hi tarah ka mobile carrier hai, jiska charge $q$ aur number density $n$ hai.

DerivationSeedhi current-carrying rod par force

Rod mein $n l A$ carriers hain. Steady current mein har ek average mein Chapter 3 ki drift velocity $𝐯_{d}$ se drift karta hai, isliye har ek par magnetic force $q 𝐯_{d} \times 𝐁$ lagta hai. Sab par lagne wale forces ko jodne par,

𝐅 = (n l A) q 𝐯_{d} \times 𝐁

Yahan $n q 𝐯_{d}$ current density $𝐣$ hai, aur $∣ 𝐣 ∣ A$ current $I$ hai. Maan lo $𝐥$ length $l$ ka vector hai jo rod ke saath current ki direction mein point karta hai. Tab $(n q 𝐯_{d}) A l = (j A) 𝐥 = I 𝐥$ , aur

Seedhe conductor par force

𝐅 = I 𝐥 \times 𝐁

(4.4)

Dekho direction kahan chali gayi. $I$ sirf ek number hai (sign ke saath), isliye current ki direction vector $𝐥$ ke paas hai. Force ka size $I l B \sin θ$ hai, jahan $θ$ rod aur field ke beech ka angle hai.

3/3Saare $n l A$ carriers par forces jud kar $𝐅 = I 𝐥 \times 𝐁$ banate hain, jo yahan upar point karta hai aur jiska size $I l B$ hai.

Fig. 4.4Drift karte carriers par lagne wale forces jud kar poori rod par

I 𝐥 \times 𝐁

force banate hain.NCERT p. 110

Ek horizontal rod, jiske dono ends leads se jude hain, crosses wale region mein padi hai jo page ke andar jaata magnetic field dikhate hain; ek label par "page ke andar" likha hai. Leads par arrowheads right ki taraf behta current I dikhate hain. Step 1: sirf rod, field aur current dikhaye gaye hain. Step 2: rod ke andar teen negative charge carriers left ki taraf drift karte hain, jo chhote blue arrows se dikhaye gaye hain, aur har ek par force q v_d cross B ke liye upar point karta chhota orange arrow hai. Step 3: chhote arrows halke pad jaate hain aur rod ke beech se ek bada orange arrow upar point karta hai, label F equals I l cross B, aur rod ke neeche ek double-headed arrow uski length l mark karta hai.

(4.4) ke saath do baatein dhyaan rakho. Pehli, $𝐁$ external field hai, woh field nahi jo rod ka apna current banata hai: wire khud ko aage nahi dhakel sakti. Doosri, yeh formula uniform field mein wire ke seedhe piece ke liye hai. Kisi bhi shape ki wire ko chhote, lagbhag seedhe pieces $d 𝐥_{j}$ mein toda ja sakta hai, aur un par forces ko vectors ki tarah joda ja sakta hai,

𝐅 = \sum_{j} I d 𝐥_{j} \times 𝐁

yeh sum integral ban jaata hai jab pieces bahut chhote kar diye jaate hain.

Try karke dekho

Wire ko uchhalo

Tumhe chahiye

ek strong disc magnet
aluminium kitchen foil ki ek patti
ek 1.5 V cell

Pehle guess karo

Jab current behta hai, to foil kis taraf hil sakti hai?

Sirf apni length ke saathSide mein, current aur field dono ke right angle parMagnet ke sabse paas wale pole ki taraf

Ek strong magnet (neodymium disc achha kaam karta hai) table par flat rakho, ek pole upar ki taraf.
Lagbhag $1 c m$ chaudi aur $15 c m$ lambi aluminium foil ki patti magnet ke across dheeli rakho taaki woh thoda latke.
Patti ke ends ko ek pal ke liye $1.5 V$ cell ke terminals par pakdo aur patti ke beech wale hisse ko dekho.
Cell ulta karo, phir magnet palto, aur dohrao.

Example 4.1NCERT Example 4.1 · p. 111

Magnetic field se tiki rod

$60 g$ mass aur $40 c m$ length ki copper rod do halki, flexible leads se horizontally latki hai. Yeh $0.35 T$ ke uniform horizontal magnetic field mein hai jo page ke andar, rod ke right angle par hai. Leads dheeli padne ke liye rod mein kitna current, aur kis direction mein, behna chahiye? $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.2NCERT Example 4.2 · p. 111

Push kis taraf hai?

Ek vacuum chamber mein magnetic field vertically upar point karta hai. Axes lo: $x$ east, $y$ north aur $z$ upar. Magnetic force ki direction nikaalo (a) east ki taraf chalte proton par, (b) east ki taraf chalte electron par, (c) east ki taraf chalte neutron par, aur (d) seedha upar chalte proton par.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

0.50 m

lambi seedhi wire mein

3.0 A

current

0.40 T

ke uniform field ke right angle par behta hai. Us par force ka size kitna hai, newtons mein?

Answer dekho

Answer 0.6 N

F = I l B = (3.0) (0.50) (0.40) = 0.60 N

4.3 Motion in a Magnetic Field

Koi force moving particle par jis rate se work karta hai, woh power $P = 𝐅 \cdot 𝐯$ hai. Magnetic force ke liye $𝐅 \cdot 𝐯 = q (𝐯 \times 𝐁) \cdot 𝐯 = 0$ , kyunki cross product un dono vectors ke perpendicular hota hai jinse woh banta hai. Isliye magnetic field charged particle par kabhi work nahi karta aur uski kinetic energy kabhi nahi badalta. Yeh particle ko mod deta hai, uske momentum ki direction badal deta hai, lekin use tez ya dheema nahi kar sakta. Particle ko energy dene ke liye electric field chahiye, jiska force $q 𝐄$ motion ke saath point kar sakta hai; har particle accelerator isi tarah energy deta hai.

Field ke right angle par velocity

Maan lo mass $m$ aur charge $q$ wala particle uniform field $𝐁$ mein speed $v$ se chalta hai, aur $𝐯$ , $𝐁$ ke perpendicular hai. Magnetic force ka size constant $q v B$ hai aur yeh hamesha side mein, motion ke right angle par point karta hai. Constant size ka force jo velocity ke perpendicular rahe, wahi hai jo uniform circular motion ko chahiye: magnetic force centripetal force deta hai, aur particle $𝐁$ ke perpendicular plane mein circle mein chalta hai (Fig. 4.5).

DerivationCircular orbit ka radius aur frequency

Radius $r$ ke circle mein speed $v$ se chalte particle ko centre ki taraf $m v^{2} / r$ ka centripetal force chahiye (Class XI, Motion in a Plane). Yahan woh force magnetic force hai, jiska size $q v B$ hai:

\frac{m v^{2}}{r} = q v B

Radius ke liye solve karne par,

Circular path ka radius

r = \frac{m v}{q B}

(4.5)

Particle ki angular speed $ω = v / r = q B / m$ hai. $ω = 2 π ν$ likhne par, jahan $ν$ per second revolutions ki sankhya hai,

Orbit ki angular frequency

ω = 2 π ν = \frac{q B}{m}

(4.6(a))

Ek revolution ka time $T = 2 π / ω = 2 π m / q B = 1 / ν$ hai.

(4.5) kehta hai ki zyada momentum $m v$ wala particle bade circle par chalta hai, aur strong field use chhote, tight circle mein mod deta hai. (4.6(a)) mein ek surprise hai: speed cancel ho gayi hai. Tez particle bade circle ka chakkar lagata hai aur dheema particle chhote circle ka, lekin dono ko bilkul utna hi time lagta hai. Frequency $ν = q B / 2 π m$ ko cyclotron frequency (साइक्लोट्रॉन आवृत्ति) kehte hain; yeh sirf particle ke charge-to-mass ratio aur field par depend karti hai.

3/5Path radius $r = m v / q B$ ke circle mein band ho jaata hai, jo time $T = 2 π m / q B$ mein poora hota hai.

Fig. 4.5

𝐯 ⊥ 𝐁

hone par magnetic force centripetal force hai aur path circle hai.

𝐁

ke saath velocity component par koi asar nahi hota, isliye circle khinch kar helix ban jaata hai.NCERT Figs. 4.5, 4.6 · p. 112

Steps 1 se 3 mein crosses wala ek square region hai, yaani page ke andar jaata magnetic field. Neeche ek positive charge right ki taraf chalta hai, blue velocity arrow ke saath aur path ke centre ki taraf upar point karte orange force arrow ke saath. Step 1: particle chalna shuru karta hai. Step 2: woh anticlockwise chauthai chakkar laga chuka hai, uska path purple mein bana hai; ab v upar aur F left point karta hai, phir se centre ki taraf. Step 3: poora circle bana hai, radius r mark hai aur paas mein formulas r equals m v over q B aur T equals 2 pi m over q B likhe hain. Steps 4 aur 5 mein side view hai jismein green field lines right ki taraf point karti hain. Step 4: charge ki velocity field se 65 degrees ka angle banati hai; charge se dashed arrows uske do components dikhate hain, field ke saath chhota v parallel aur uske across lamba v perpendicular. Step 5: path ek purple helix hai jo field ke saath ek dashed axis ke around lipta hai, saamne ke hisse solid aur peeche ke dashed; do turns ke beech pitch p mark hai aur right end par axis se radius r mark hai, label p equals v parallel T ke saath.

Angle par velocity: helix

Ab maan lo $𝐯$ , $𝐁$ se angle $θ$ banata hai. Velocity ko field ke saath component $v_{∥} = v \cos θ$ aur uske across component $v_{⊥} = v \sin θ$ mein todo. Magnetic force sirf $v_{⊥}$ par depend karta hai ( $𝐁$ ka $𝐁$ ke saath wale vector se cross product zero hota hai), isliye $v_{∥}$ kabhi nahi badalta. Perpendicular motion pehle ki tarah radius $m v_{⊥} / q B$ ka circle hai. Dono ko milao, to particle field ke around ghoomte hue uske saath barabar aage badhta hai: uska path ek helix (कुंडलिनी) hai, screw ki chudi jaisa.

Helix ko do numbers describe karte hain. Uska radius, $r = m v_{⊥} / q B$ , us circle ka radius hai jo tumhe field ke saath seedha dekhne par dikhega. Uski pitch $p$ (चूड़ी अंतराल) yeh hai ki particle ek turn mein field ke saath kitna aage badhta hai, jaise screw ki chudiyon ke beech ki doori:

Helix ki pitch

p = v_{∥} T = \frac{2 π m v_{∥}}{q B}

(4.6(b))

InteractiveMagnetic field mein charged particle

Ek interactive stage jo uniform magnetic field mein chalta charged particle dikhata hai. Circle view mein field page ke andar ya bahar point karta hai, jo crosses ya dots se dikhaya gaya hai, aur particle circle par chalta hai, uski velocity aur centre ki taraf magnetic force ke arrows ke saath; ek button same point se aadhi speed par chalne wala twin particle jodta hai. Helix view mein field right ki taraf point karta hai aur particle uske saath spiral karta hai; helix ka saamne aur peeche ka hissa solid aur dashed lines se dikhaya gaya hai aur pitch mark hai. Tum proton, alpha particle ya electron chun sakte ho, aur sliders se speed, field strength aur velocity aur field ke beech ka angle set kar sakte ho. Readouts radius, period, frequency, pitch, kinetic energy aur screen par ek chakkar ka time dikhate hain. 5 cm ka scale bar size set karta hai.

Fig. 4.6Ek particle chuno, phir uski speed, field aur dono ke beech ka angle badlo. Readouts radius, period, pitch aur kinetic energy dikhate hain; motion slow motion mein dikhaya gaya hai.

Try karke dekho

Orbit simulation ke saath try karne wali cheezein

Circle view mein radius aur screen par ek chakkar ka time note karo. Speed dugni karo. Dono mein se kya badalta hai, aur kitna?
Add a twin at half speed dabao. Twin aadhe size ke circle ka chakkar lagata hai. Dono particles ko dekho: woh phir kahan aur kab milte hain?
Speed $2.0 \times 10^{6} m / s$ aur field $0.40 T$ rakho, aur proton se alpha particle par switch karo. Radius, chakkar ka time aur kinetic energy compare karo.
Electron par switch karo aur corner mein rotation label padho. Phir field ko page se bahar kar do. Rotation ki direction charge ke sign aur $𝐁$ ki direction par kaise depend karti hai?
Helix view mein angle ko $90^{\circ}$ se $20^{\circ}$ tak slide karo. Radius, pitch aur period ka kya hota hai?

$v$ dugna karne se $r$ dugna hota hai lekin chakkar ka time same rehta hai, aur twin har chakkar ke baad starting point par original particle se milta hai: period $2 π m / q B$ speed par depend nahi karta. Alpha particle ka $m / q$ proton se dugna hai, isliye same speed par uska radius aur period dono dugne hote hain, jabki uski kinetic energy chaar guna hoti hai kyunki uska mass proton ka chaar guna hai. Electron proton ke ulta ghoomta hai, aur field palatne se dono ulte ho jaate hain. $𝐯$ ko $𝐁$ ki taraf jhukane se radius chhota aur pitch lambi hoti hai, jabki period same rehta hai.

Example 4.3NCERT Example 4.3 · p. 113

Proton ka track padhna

Ek beam line mein $0.25 T$ ke uniform field wala bending magnet protons (mass $1.67 \times 10^{- 27} k g$ , charge $1.6 \times 10^{- 19} C$ ) ko field ke right angle par radius $10.0 c m$ ke circular arc par modta hai. Nikaalo (a) protons ki speed, (b) unki kinetic energy keV mein, $1 e V = 1.6 \times 10^{- 19} J$ lekar, aur (c) proton ko poore circle ka ek chakkar lagane mein kitna time lagega, aur woh per second kitne chakkar lagayega.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Poles ke upar spiral karte electrons

Polar regions ke bahut upar, jahan Earth ka field lagbhag $5.0 \times 10^{- 5} T$ hai, ek electron ( $m = 9.1 \times 10^{- 31} k g$ ) field se $60^{\circ}$ banane wali direction mein $1.0 \times 10^{7} m / s$ se chalta hai. Uske helical path ka radius aur pitch, aur ek turn mein lagne wala time nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek proton uniform magnetic field mein ghoom raha hai. Field same rakhte hue uski speed dugni kar di jaati hai. Ek revolution ke time ka kya hota hai?

Aadha ho jaata haiSame rehta haiDugna ho jaata hai

Answer dekho

Answer (B) Same rehta hai

T = 2 π m / q B

mein

v

nahi hai. Radius dugna ho jaata hai, isliye particle dugni speed se dugni doori same time mein tay karta hai.

4.4 Magnetic Field due to a Current Element, Biot–Savart Law

Ab tak magnetic field bas diya hua tha. Ab hum poochte hain ki current use kaise banata hai. Magnetic field ke sirf do sources pata hain: electric currents, yaani moving charges, aur electrons jaise particles ka apna intrinsic magnetism. Yeh section pehle wale ke baare mein hai. Currents ka rule 1820 mein Jean-Baptiste Biot aur Félix Savart ne dhoondha, jinhone current-carrying wires se alag-alag doori par magnetised needle ke swings ka time naapa.

Socho ek wire jismein steady current $I$ hai, aur use mann mein chhote-chhote pieces mein kaat lo. Length $d l$ ka ek piece chuno, aur $d 𝐥$ us length ka vector ho jo current ke saath point kare. Product $I d 𝐥$ ko current element kehte hain. Humein woh field $d 𝐁$ chahiye jo yeh element point P par banata hai, jahan $𝐫$ element se P tak ka displacement hai aur $θ$ , $d 𝐥$ aur $𝐫$ ke beech ka angle hai (Fig. 4.7).

LawBiot–Savart law

Vacuum mein current element $I d 𝐥$ kisi point par jo field $d 𝐁$ banata hai, woh $I$ aur $d l$ ke proportional badhta hai, element se doori $r$ ke saath $1 / r^{2}$ ki tarah ghatta hai, aur $d 𝐥$ aur $𝐫$ ke plane ke right angle par, $d 𝐥 \times 𝐫$ ke saath point karta hai.

Symbols mein,

Biot–Savart law

d 𝐁 = \frac{μ_{0}}{4 π} \frac{I d 𝐥 \times 𝐫}{r^{3}}

(4.7(a))

Kyunki $∣ d 𝐥 \times 𝐫 ∣ = r d l \sin θ$ , field ka magnitude hai

d B = \frac{μ_{0}}{4 π} \frac{I d l \sin θ}{r^{2}}

(4.7(b))

Constant $μ_{0}$ ko permeability of free space (निर्वात की चुंबकशीलता) kehte hain. SI units mein iski value hai

\frac{μ_{0}}{4 π} = 10^{- 7} T m / A

(4.7(c))

isliye $μ_{0} = 4 π \times 10^{- 7} T m / A$ .Kai dashakon tak yeh value definition se exact thi. 2019 ke SI revision ke baad $μ_{0}$ ek measured quantity hai, lekin yeh ab bhi ek billion mein ek part se bhi behtar accuracy tak $4 π \times 10^{- 7} T m / A$ ke barabar hai. Direction ke liye cross product ka right-hand rule use karo: ungliyan $d 𝐥$ ke saath, $𝐫$ ki taraf mudi hui, aur thumb $d 𝐁$ deta hai. Fig. 4.7 mein isse upar ki taraf current element ke right wale points par field page ke andar aur left wale points par page se bahar aata hai.

1/4Element ke right angle par $\sin θ = 1$ aur $d B$ sabse bada hai. $d 𝐥 \times 𝐫$ page ke andar point karta hai.

Fig. 4.7Current element ka field uske right angle par sabse bada, uski line ke saath zero, aur ulti side par ulti direction mein hota hai.NCERT Fig. 4.7 · p. 113

Beech mein ek chhota orange arrow upar point karta hai, label I dl, aur uske through ek dashed vertical line element ki line mark karti hai. Right mein ek bar gauge dB ka size dikhata hai, aur top left par Biot–Savart formula likha hai. Step 1: P element ke theek right mein doori r par hai, theta 90 degrees; P par field ek bade purple cross se dikhaya gaya hai, page ke andar, aur gauge bhara hua hai. Step 2: P same doori par ghoom kar horizontal se 45 degrees upar aa gaya hai; cross chhota hai aur gauge lagbhag saat-das bhara hai. Step 3: P element ke theek upar dashed line par hai; koi field nahi hai, dB equals 0 mark hai, aur gauge khaali hai. Step 4: P element ke theek left mein hai; field ek bada purple dot hai, page se bahar, aur gauge phir bhara hua hai.

Biot–Savart law ko Coulomb's law ke saath rakh kar dekhna faydemand hai, jo point charge ka electric field deta hai (Table 4.3).

Table 4.3Coulomb's law aur Biot–Savart law ki tulnaNCERT p. 114

Feature	Point charge ka electric field	Current element ka magnetic field
Source	charge $q$ , ek scalar	current element $I d 𝐥$ , ek vector
Doori par dependence	$1 / r^{2}$	$1 / r^{2}$
Direction	$𝐫$ ke saath, charge se door ya uski taraf	$d 𝐥$ aur $𝐫$ ke plane ke right angle par
Angle par dependence	koi nahi: har direction mein same	$\sin θ$ ; element ki line ke saath zero
Kai sources	fields vectors ki tarah judte hain	fields vectors ki tarah judte hain
Constant	$1 / 4 π ε_{0} = 9 \times 10^{9} N m^{2} / C^{2}$	$μ_{0} / 4 π = 10^{- 7} T m / A$

Electricity aur magnetism ke dono constants vacuum mein speed of light $c$ se jude hain:

μ_{0} ε_{0} = (4 π ε_{0}) (\frac{μ_{0}}{4 π}) = \frac{1}{9 \times 10^{9}} \times 10^{- 7} = \frac{1}{(3 \times 10^{8})^{2}} = \frac{1}{c^{2}}

Yeh match koi sanyog nahi hai. Light ek electromagnetic wave hai, aur Chapter 8 dikhata hai ki uski speed in do constants se tay hoti hai, $c = 1 / \sqrt{μ_{0} ε_{0}}$ . Kyunki $c$ exactly pata hai, $μ_{0}$ aur $ε_{0}$ same information rakhte hain: jo laboratory inmein se ek ko pakka naap leti hai, usne asal mein doosre ko bhi naap liya.

Example 4.4NCERT Example 4.4 · p. 114

Wire ke chhote piece ka field

$2.0 m m$ lamba wire ka ek chhota seedha piece origin par $z$ -axis ke saath pada hai aur usmein $+ z$ direction mein $15 A$ current hai. Iska banaya field nikaalo (a) $x$ -axis par point $(0.30 m, 0, 0)$ par, (b) $z$ -axis par $(0, 0, 0.30 m)$ par, aur (c) $x z$ -plane mein origin se $0.30 m$ door us point par jo $+ z$ axis se $30^{\circ}$ banane wali line par hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek current element north point karta hai. Usse same doori par kis point par uska field zero hai?

Uske theek north wala pointUske theek east wala pointUske theek upar wala point

Answer dekho

Answer (A) Uske theek north wala point

Theek north element ki line ke saath hai, jahan

θ = 0

aur

\sin θ = 0

. East aur upar dono element se

90^{\circ}

par hain, jahan field sabse bada hai.

4.5 Magnetic Field on the Axis of a Circular Current Loop

Biot–Savart law kisi bhi current ka field de deta hai, jab uske saare elements ke contributions jod diye jaayein. Sabse simple shape jiske liye yeh sum exactly kiya ja sakta hai, circular loop hai, aur ise karne ki sabse simple jagah loop ke axis par ek point hai.

Vacuum mein radius $R$ ka ek loop lo jismein steady current $I$ hai, uska centre O origin par, uska plane $y z$ -plane mein aur uska axis $x$ -axis ke saath (Fig. 4.8). Humein axis par O se $x$ doori wale point P par field chahiye.

3/4Theek saamne (diametrically opposite) wala element ulti direction mein barabar $d B_{⊥}$ deta hai, jo cancel ho jaata hai, aur barabar $d B_{x}$ , jo jud jaata hai.

Fig. 4.8Loop ke aamne-saamne ke elements axis ke right angle wale field components ko ek doosre se cancel kar dete hain, isliye sirf axial components judte hain.NCERT Fig. 4.9 · p. 115

Left mein ek circular loop lambe ellipse ki tarah bana hai, uska saamne ka aadha solid aur peeche ka aadha dashed, aur arrows current dikhate hain. Loop ke top par dot hai, current page se bahar, aur bottom par cross, current page ke andar. Ek horizontal axis centre O se guzar kar right mein doori x par point P tak jaata hai; O se top tak radius R mark hai. Step 1: ek line r top element d l ko P se jodti hai, aur P par ek purple arrow d B, r ke right angle par, upar aur right ki taraf point karta bana hai. Step 2: d B ko axis ke saath component d B x aur upar point karte component d B perpendicular mein toda gaya hai, d B aur axis ke beech angle theta ke saath. Step 3: bottom element d l prime ek field d B prime deta hai jo neeche aur right ki taraf point karta hai; iska perpendicular component neeche point karke upar wale ko cancel karta hai, jabki dono axial components right ki taraf point karte hain. Step 4: baaki arrows halke pad jaate hain aur P par axis ke saath ek mota arrow B point karta hai, formula B equals mu nought I R squared over 2 times x squared plus R squared to the power three halves ke saath.

DerivationCurrent ki ring ke axial point par field

Loop ka har element $d 𝐥$ , P se same doori $r = \sqrt{x^{2} + R^{2}}$ par hai. Har element use P se jodne wali line ke perpendicular bhi hai: circle ka tangent element axis aur radius dono ke right angle par hai, aur isliye unke plane ke bhi. Isliye $∣ d 𝐥 \times 𝐫 ∣ = r d l$ , aur (4.7(a)) se ek element ke field ka size hai

d B = \frac{μ_{0}}{4 π} \frac{I ∣ d 𝐥 \times 𝐫 ∣}{r^{3}}

(4.8)

d B = \frac{μ_{0}}{4 π} \frac{I d l}{x^{2} + R^{2}}

(4.9)

$d 𝐁$ , $d 𝐥$ aur $𝐫$ dono ke right angle par hai, isliye woh axis se jhuka hua hai. Iska ek component $d B_{x}$ axis ke saath hai aur ek component $d B_{⊥}$ axis ke right angle par.

Har element ko uske theek saamne wale element ke saath jodi banao. Unke perpendicular components barabar aur ulte hain aur cancel ho jaate hain, jabki axial components same taraf point karke jud jaate hain. Isliye loop par sum karne par sirf $d B_{x} = d B \cos θ$ bachta hai, jahan $θ$ , $d 𝐁$ aur axis ke beech ka angle hai. Figure ki geometry se,

\cos θ = \frac{R}{(x^{2} + R^{2})^{1 / 2}}

(4.10)

aur isliye

d B_{x} = \frac{μ_{0} I d l}{4 π} \frac{R}{(x^{2} + R^{2})^{3 / 2}}

Is expression mein $d l$ ke alawa sab kuch har element ke liye same hai. Lengths $d l$ jud kar circumference $2 π R$ banti hain, isliye P par field hai

Loop ke axis par field

𝐁 = B_{x} \hat{𝐢} = \frac{μ_{0} I R^{2}}{2 (x^{2} + R^{2})^{3 / 2}} \hat{𝐢}

(4.11)

Loop ke centre par $x = 0$ , aur field hai

Loop ke centre par field

𝐁_{0} = \frac{μ_{0} I}{2 R} \hat{𝐢}

(4.12)

$N$ paas-paas lapete turns wali flat coil ek ke upar ek rakhe $N$ loops ki tarah kaam karti hai, isliye dono results $N$ se multiply ho jaate hain. Axis par bahut door, jahan $x ≫ R$ , field $1 / x^{3}$ ki tarah ghatta hai; is par hum Section 4.9.2 mein wapas aayenge.

Current loop ki field lines closed curves hain jo loop ke beech se guzarti hain (Fig. 4.9). Unki direction right-hand thumb rule se milti hai: right hand ki ungliyan loop ke around current ki direction mein modo, to thumb loop ke andar field ki direction dikhata hai. Jis face se field lines bahar nikalti hain, wahan se dekhne par current anticlockwise behta hai. Woh face magnet ke north pole ki tarah behave karta hai, aur ulta face, jahan current clockwise dikhta hai, south pole ki tarah.

3/3Field lines ek face se nikalti hain aur doosre mein ghusti hain, bilkul jaise bar magnet ke N pole se nikal kar S pole mein ghusti hain.

Fig. 4.9Current loop ki field lines closed loops hain. Jis face se woh nikalti hain, woh north pole ki tarah kaam karta hai.NCERT Fig. 4.10 · p. 116

Step 1: ek circular loop khade ellipse ki tarah bana hai, saamne ka aadha solid aur peeche ka aadha dashed, arrows current dikhate hain aur ek dashed horizontal line uska axis mark karti hai. Step 2: loop halka pad jaata hai aur uske kate hue ends dikhte hain, top par dot jahan current page se bahar aata hai aur bottom par cross jahan woh andar jaata hai. Blue field lines har kate end ke around ghoomti hain, aur axis ke paas ki lines loop ke beech se left se right seedhi guzarti hain, right mein phail jaati hain aur left mein mud kar wapas aati hain; axis ke saath ek seedhi line hai, label B. Step 3: loop ke left face par S aur right face par N likha aata hai, aur corner mein S aur N label wala ek chhota bar magnet hai jiske paas likha hai "magnet jaisa".

Example 4.5NCERT Example 4.5 · p. 117

Pie ka ek slice

Wire ka ek loop pie ke slice jaisa hai: ek circle ke radii ke saath do seedhi sides aur radius $4.0 c m$ ka circular arc jo centre O par $90^{\circ}$ banata hai (Fig. 4.10). Ismein $10 A$ current behta hai, figure mein dekhne par anticlockwise. (a) O par magnetic field ka size aur direction nikaalo. (b) Same radius aur current rakhte hue, O par field $5.0 \times 10^{- 5} T$ , yaani lagbhag Earth ke field jitna, hone ke liye arc ko kitna angle banana hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 4.10Example 4.5 ka loop. Sirf curved side O par field banati hai.NCERT Fig. 4.11 · p. 117

Pie ke slice jaisa loop jiske nok O par, neeche left, right angle hai. Ek seedhi side radius R ke saath right ki taraf jaati hai, curved side quarter circle hai, aur doosri seedhi side neeche O tak wapas aati hai. Arrowheads dikhate hain ki current I neeche wali side par right ki taraf, arc par anticlockwise upar aur ghoom kar, aur left side par neeche O tak behta hai. O par 90 degree ka angle mark hai.

Example 4.6NCERT Example 4.6 · p. 117

Earth ke field ko cancel karna

Physics laboratory ko aisa chhota region chahiye jahan koi magnetic field na ho. 250 turns aur mean radius $5.0 c m$ wali flat circular coil ke centre par Earth ke field ka horizontal component, north ki taraf $3.5 \times 10^{- 5} T$ , exactly cancel karna hai. Kitna current chahiye, aur coil ko kaise rakhna hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

50 turns aur radius

5.0 c m

wali flat coil mein

2.0 A

current hai. Uske centre par field kitna hai, millitesla mein?

Answer dekho

Answer 1.2566 mT

B = μ_{0} N I / 2 R = (4 π \times 10^{- 7}) (50) (2.0) / (0.10) = 1.26 \times 10^{- 3} T = 1.26 m T

4.6 Ampère's Circuital Law

Biot–Savart law se current elements jodna hamesha kaam karta hai, lekin integrals mushkil ho sakte hain. Ampère's circuital law wahi physics alag tarah se pack karta hai, aur jab bhi current ka arrangement kaafi symmetric ho, yeh field lagbhag muft mein de deta hai. Magnetism ke liye yeh wahi karta hai jo electrostatics ke liye Gauss's law karta hai.

Yeh law ek closed path C ke around $𝐁$ ke circulation naam ki quantity ke baare mein hai. Socho tum ek probe ko C ke around ek baar ghuma rahe ho, aur ginte ja rahe ho ki field tumhari chalne ki direction mein kitna point karta hai: har chhote step $d 𝐥$ ke liye tum $𝐁 \cdot d 𝐥$ jodte ho, jo positive hai jahan field tumhari jaane ki taraf point kare, negative jahan peeche point kare, aur zero jahan tumhare path ke across point kare. Ek poore chakkar ka total $\oint_{C} 𝐁 \cdot d 𝐥$ likha jaata hai (Fig. 4.11). Ampère's law kehta hai ki yeh total sirf us current par depend karta hai jo path ke beech se guzarta hai.

LawAmpère's circuital law

Kisi bhi closed curve ke around magnetic field ka line integral $μ_{0}$ times us net current ke barabar hota hai jo us curve se ghiri kisi bhi surface se guzarta hai.

Ampère's circuital law

\oint_{C} 𝐁 \cdot d 𝐥 = μ_{0} I

(4.13(a))

jahan

$I$: C se ghiri surface se guzarne wala net current, right-hand rule se mile sign ke saath gina hua

3/4Right hand ki ungliyan C ke saath modo: thumb page se bahar point karta hai, isliye $I_{1}$ positive count hota hai aur $I_{2}$ negative. $I_{3}$ surface ko nahi cheerta aur count nahi hota.

Fig. 4.11Ampère's law closed curve ke around

𝐁 \cdot d 𝐥

jodta hai. Sirf woh currents count hote hain jo curve ko cheerte hain, aur unke signs right-hand rule se tay hote hain.NCERT Fig. 4.12 · p. 117

Teen wires page ko right angle par cross karti hain: I1, ek dot, aur I2, ek cross, anticlockwise bane green closed curve C ke andar hain, aur I3, ek dot, uske bahar hai. Step 1: curve aur teen currents. Step 2: C ki right side ke ek point par chhota black arrow d l curve ke saath point karta hai, purple arrow B upar aur right ki taraf point karta hai, aur dashed purple line B t dikhati hai, yaani B ka d l ke saath wala part. Step 3: labels badal kar plus I1 aur minus I2 ho jaate hain, I3 ko bahar mark kiya gaya hai, aur equation line integral of B dot d l equals mu nought times I1 minus I2 dikhayi gayi hai. Step 4: curve ki jagah radius r ka circle aa jaata hai jiska centre page se bahar current I wali single wire hai; chaar points par circle ke saath anticlockwise point karte B ke purple arrows hain, aur equation B times 2 pi r equals mu nought I.

Kaun se currents positive gine jaate hain? C ko us side se dekho jahan se uske around tumhara chalna anticlockwise dikhe. Tumhari taraf aate currents positive gine jaate hain aur tumse door jaate negative. (Yeh right-hand rule hi hai: ungliyan chalne ke saath, thumb positive current ke saath.) Jo currents C ke beech se nahi guzarte, woh bilkul count nahi hote, chahe uske kitne bhi paas se guzrein; woh path par $𝐁$ badalte hain, lekin circulation mein unke contributions jud kar zero ho jaate hain.

Circulation nikaalna aasaan hai jab path field ki symmetry follow kare. Jis hisse par $𝐁$ path ke saath constant size $B$ se chale, wahan contribution bas $B$ times us hisse ki length hai. Jin hisson par $𝐁$ path ko right angle par cross kare, ya zero ho, wahan contribution zero hai. Sirf aise hisson se bana path Amperian loop (ऐम्पियरी पाश) kehlata hai. Agar pehli tarah ke hisson ki total length $L$ ho aur loop net current $I_{e}$ ko ghere, to law ban jaata hai

Achhe se chune loop ke liye Ampère's law

B L = μ_{0} I_{e}

(4.13(b))

Long straight wire ka field

Current $I$ wali infinitely long straight wire mein cylindrical symmetry hai. Wire ko uske apne axis ke around ghumane ya uski length ke saath khiskane par kuch nahi badalta. Isliye field sirf wire se doori $r$ par depend kar sakta hai, aur woh wire ke around ghoomta hai. Amperian loop ke roop mein wire par centred radius $r$ ka circle chuno, wire ke right angle wale plane mein. $𝐁$ is circle ka tangent hai aur poore circle par same size ka hai, isliye $L = 2 π r$ aur $I_{e} = I$ :

B (2 π r) = μ_{0} I

Long straight wire ka field

B = \frac{μ_{0} I}{2 π r}

(4.14)

Teen quick checks is result ko maanne layak banate hain. Doori: field $1 / r$ ki tarah ghatta hai, single current element ke $1 / r^{2}$ se dheere, kyunki lambi wire ke elements tumse har doori par hain. Infinity: wire kabhi khatam nahi hoti, phir bhi wire ke bahar har jagah field finite hai; uske door ke hisse door hain aur lagbhag end-on dikhte hain, jahan $\sin θ$ chhota hai, isliye woh kam se kam judte hain. Size: single seedhi wire mein $10 A$ current $10 c m$ par $2 \times 10^{- 5} T$ deta hai. Yeh Earth ke field ka lagbhag aadha hai, laboratory ke hisaab se weak, phir bhi itne paas rakhe compass ko north se lagbhag $30^{\circ}$ hata dene ke liye kaafi.

Kyunki wire par centred kisi bhi circle par field ka size same hai aur woh circle ke saath point karta hai, field lines wire ke around rings hain, bilkul wahi pattern jo Oersted ke compasses ne banaya tha. Har magnetic field line ki tarah yeh apne aap mein band hoti hain, na shuru na khatam; iske ulta electric field lines charges par shuru aur khatam hoti hain. Right-hand grip rule inki direction deta hai: right hand ko wire ke around lapeto aur thumb current ke saath point karo, to ungliyon ka mudna batata hai ki field lines kis taraf chalti hain.

Kab kaun sa law use karein? Biot–Savart law hamesha answer deta hai, bhale hi mushkil integral karna pade. Ampère's law hamesha sach hai, lekin $𝐁$ nikaalne ke liye kaam ka tabhi hai jab symmetry humein $B$ ko integral se bahar nikaalne de: long straight wires, thick wires, solenoids, toroids. Single circular loop ke liye kisi path par field constant nahi hai, isliye centre par field Section 4.5 mein Biot–Savart law se hi nikaalna pada. Dono laws rivals nahi hain: Ampère's law ko Biot–Savart law se prove kiya ja sakta hai, aur steady currents ke liye dono mein same physics hai.

Example

Solar cable aur compass

Ek single seedhi cable rooftop solar array se inverter tak $20 A$ ka steady current le jaati hai. Compass ko cable se kitni door hona chahiye taaki cable ka field ghat kar $5 \times 10^{- 5} T$ , yaani lagbhag Earth ke field jitna, ho jaaye? Isse paas compass reading ka kya hota hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

DerivationThick wire ke andar aur bahar field

Ab wire ko mota karo: steady current $I$ ko radius $a$ ke solid cylinder se behne do, jiske cross-section mein har jagah same current density ho. Symmetry se field axis ke around ghoomta hai, aur uska size sirf axis se doori $r$ par depend karta hai. Axis par centred radius $r$ ka circular Amperian loop use karo.

Wire ke bahar ( $r > a$ ) loop poore current ko gherta hai, $I_{e} = I$ , aur humein patli wire ka field wapas milta hai:

B = \frac{μ_{0} I}{2 π r} (r > a)

(4.15(a))

Wire ke andar ( $r < a$ ) loop sirf area $π r^{2}$ se behne wale current ko gherta hai. Current density uniform hai, isliye

I_{e} = I \frac{π r^{2}}{π a^{2}} = \frac{I r^{2}}{a^{2}}

Phir Ampère's law se $B (2 π r) = μ_{0} I r^{2} / a^{2}$ , isliye

B = \frac{μ_{0} I}{2 π a^{2}} r (r < a)

(4.15(b))

Wire ke andar field $r$ ke proportional badhta hai, surface par apni sabse badi value $μ_{0} I / 2 π a$ tak pahunchta hai, aur bahar $1 / r$ ki tarah ghatta hai.

Example 4.7NCERT Example 4.7 · p. 119

Coaxial cable ke andar

Coaxial cable, jaisi ghar mein TV ya broadband signal laati hai, mein radius $a = 1.0 m m$ ki solid inner wire aur same axis wali radius $b = 4.0 m m$ ki patli cylindrical outer sheath hoti hai. $2.0 A$ ka steady current inner wire mein, uske cross-section par barabar phaila hua, behta hai aur sheath se wapas aata hai. Axis se (a) $0.50 m m$ , (b) $1.0 m m$ , (c) $3.0 m m$ aur (d) $6.0 m m$ ki doori par magnetic field nikaalo, aur $B$ vs $r$ ka sketch banao (Fig. 4.12).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 4.12Coaxial cable ka field: inner wire ke andar

r

ke proportional, conductors ke beech

1 / r

ke proportional, aur bahar zero.NCERT Figs. 4.13, 4.14 · p. 120

Left mein coaxial cable ka cross-section: radius a ki chhoti solid inner wire jismein current page se bahar hai, dot se mark, aur uske chaaron taraf radius b ki moti outer ring jismein current page ke andar hai, crosses se mark. Right mein B vs r ka graph: B axis par zero se seedhi line mein r equals a par peak tak badhta hai, label B proportional to r, phir B proportional to 1 over r label wale curve ke saath r equals b tak ghatta hai, jahan woh zero par gir kar zero hi rehta hai, label B equals 0.

Quick check

4 mein se 1

Ek long straight wire mein current

20 A

hai.

10 c m

door uske magnetic field ka size kitna hai, tesla mein?

Answer dekho

Answer 0.00004 T

B = μ_{0} I / 2 π r = (2 \times 10^{- 7}) (20) / 0.10 = 4.0 \times 10^{- 5} T

4.7 The Solenoid

Maan lo kisi experiment ko aisa magnetic field chahiye jo strong, steady aur tumhari mutthi jitne region mein har jagah same ho. Single loop yeh nahi de sakta: uska field wire ke paas sabse strong hota hai aur point-point par badalta hai. Hal yeh hai ki ek common axis ke saath kai loops ek ke baad ek rakho, ek lambi wire ko tube par turn ke baad turn lapet kar, jaise reel par dhaaga. Jo coil apni chaudai se kai guna lambi ho, use solenoid (परिनालिका) kehte hain. Wire par enamel ki patli parat hoti hai, isliye paas ke turns chhoo sakte hain aur current ek turn se agle turn mein short-cut nahi karta.

Fig. 4.13 solenoid ke axis ke saath ek slice dikhata hai: upar dots ki row aur neeche crosses ki row wahi turns hain, kate hue. Har wire ke paas field uske around ghoomta hai, jaise seedhi wire ke around. Do paas ke turns ke beech unke fields ulti directions mein point karte hain aur cancel ho jaate hain. Solenoid ke andar saare turns ke fields axis ke saath same taraf point karte hain aur jud kar strong field banate hain. Bahar har turn ki paas aur door wali sides ke contributions zyadatar cancel ho jaate hain, aur field weak hota hai.

3/4Long solenoid ke liye andar field uniform aur axis ke saath hai, aur bahar field lagbhag zero hai.

Fig. 4.13Paas ke turns ke fields turns ke beech aur bahar cancel ho jaate hain, aur andar jud kar strong, uniform field banate hain.NCERT Figs. 4.15, 4.16 · p. 121

Solenoid ke chhah turns uske axis ke saath kate hue hain: upar dots ki row, currents page se bahar, aur neeche crosses ki row, currents page ke andar; dono ke beech solenoid ki dashed outline hai aur top left par current I likha hai. Step 1: har wire ke around field ka chhota blue circle hai, dots ke around anticlockwise aur crosses ke around clockwise. Step 2: upar ki do paas ki wires ke beech do chhote orange arrows ulti directions mein point karte hain, label cancel, jabki solenoid ke andar ek green arrow right point karta hai, label add. Step 3: chhote circles gayab hain aur paanch seedhi blue field lines andar se right ki taraf chalti hain, label B, aur bahar ek note: "bahar: bahut kam field". Step 4: ek dashed green rectangle a b c d joda gaya hai, jiski length h wali side a b solenoid ke andar field ke saath hai aur side c d bahar neeche wali row ke neeche hai, jo do turns gherta hai, aur equation B h equals mu nought times n h times I.

Finite length ke asli solenoid mein kuch field ends se bahar nikal jaata hai, aur bahar ka field zero nahi balki chhota hota hai. Solenoid jitna lamba, yeh leak utna kam matter karta hai. Isliye long solenoid ke liye hum simple model use karte hain: andar axis ke saath uniform field, aur bahar bilkul field nahi.

DerivationLong solenoid ke andar field

Maan lo solenoid mein per unit length $n$ turns hain, har ek mein current $I$ . Fig. 4.13 ka rectangular Amperian loop abcd chuno: length $h$ wali side ab solenoid ke andar axis ke parallel chalti hai, aur side cd uske bahar.

ab ke saath field tangential hai aur size $B$ hai, isliye yeh side $B h$ contribute karti hai. cd ke saath field zero hai. bc aur da ke saath, jahan woh solenoid ke andar hain wahan field path ke perpendicular hai aur bahar zero, isliye yeh sides kuch contribute nahi karti. Isliye $L = h$ .

Loop $n h$ turns gherta hai, har ek mein $I$ , isliye $I_{e} = n h I$ . (4.13(b)) wali form mein Ampère's law deta hai

B h = μ_{0} (n h) I

Long solenoid ke andar field

B = μ_{0} n I

(4.16)

Dekho (4.16) mein kya nahi hai. Koi radius nahi hai: per metre same turns aur same current wali patli coil aur moti coil same field deti hain. Aur koi position nahi hai: ends se door, tube ki deewar ke paas field utna hi strong hai jitna axis par. Axis ke saath iski direction loop wale right-hand rule se milti hai. Bade volume mein strong aur uniform, dono tarah ka field bilkul wahi hai jo MRI scanners aur electron-beam experiments ko chahiye, isliye solenoid magnet design ka workhorse hai. Coil mein soft iron ki rod daalne se field kai guna badh jaata hai; Chapter 5 samjhaata hai ki iron kaise saath deta hai.

Example 4.8NCERT Example 4.8 · p. 122

Solenoid design karna

Ek student ko $30 c m$ lambe solenoid ke andar $5.0 m T$ ka uniform field chahiye, aisi insulated copper wire se jo safely $2.0 A$ le ja sake. (a) Kitne turns chahiye? (b) Agar saare turns ek hi layer mein saath-saath fit hone chahiye, to wire ka maximum diameter kitna ho sakta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek long solenoid mein per metre 2000 turns hain aur usmein

0.50 A

current hai. Uske andar field kitna hai, millitesla mein?

Answer dekho

Answer 1.2566 mT

B = μ_{0} n I = (4 π \times 10^{- 7}) (2000) (0.50) = 1.26 \times 10^{- 3} T = 1.26 m T

4.8 Force between Two Parallel Currents, the Ampere

1905 mein Sydney ke do physicists ne New South Wales ki ek coal mine ke lightning conductor se nikali copper tube ke baare mein likha. Bijli girne ke baad tube ka ek hissa aise pichak gaya tha jaise vice mein dabaya gaya ho, jabki bahar se use kisi cheez ne nahi maara tha. Zimmedar tha khud lightning current, hazaaron amperes jo tube ki deewar ki har strip mein beh rahe the, aur har strip baaki ko kheench rahi thi. Hum yeh pehle se predict kar sakte hain: current ki har strip baaki ke magnetic field mein hai, aur magnetic field current ko push karta hai. Sabse saaf case do long, straight, parallel wires ka hai.

DerivationDo long parallel wires ke beech force

Maan lo wires a aur b ek doosre se $d$ doori par hain aur unmein same direction mein steady currents $I_{a}$ aur $I_{b}$ hain (Fig. 4.14). (4.14) se wire a, wire b ke har point par same size ka field banati hai:

B_{a} = \frac{μ_{0} I_{a}}{2 π d}

Right-hand grip rule dikhata hai ki yeh field wire b ke perpendicular hai. Isliye (4.4) se wire b ki length $L$ par $I_{b} L B_{a}$ size ka force lagta hai, wire a ki taraf:

Wire b ki length L par force

F_{b a} = I_{b} L B_{a} = \frac{μ_{0} I_{a} I_{b}}{2 π d} L

(4.17)

Roles badal kar yahi argument dohrane par wire a ki length $L$ par same size ka force milta hai, wire b ki taraf:

𝐅_{b a} = - 𝐅_{a b}

(4.18)

3/4Wire a par b ki taraf barabar aur ulta force lagta hai, jaisa Newton's third law kehta hai.

Fig. 4.14Parallel currents attract karte hain; ek current ulta karo to woh repel karte hain.NCERT Fig. 4.17 · p. 122

Do wires page ko right angle par cross karti hain, left mein a aur right mein b, ek doosre se doori d par, dono mein current page se bahar, dots se mark aur label I a aur I b. Step 1: dashed blue circles wire a ka field uske around anticlockwise ghoomta dikhate hain, aur wire b par label B a wala blue arrow seedha upar point karta hai. Step 2: b par ek orange arrow left, a ki taraf, point karta hai, label F b a. Step 3: a par ek barabar orange arrow right, b ki taraf, point karta hai, label F a b, aur formula f equals mu nought I a I b over 2 pi d dikhaya gaya hai. Step 4: b mein current ulta hai, cross se bana, aur dono force arrows ab bahar ki taraf, ek doosre se door, point karte hain, note "dono repel" ke saath.

Wire b, a ki taraf kheenchi jaati hai, aur wire a, b ki taraf, barabar size ke forces se: steady currents ke liye yeh pair Newton's third law follow karta hai. Ab b mein current ulta karo (Fig. 4.14, last step). b par a ka field nahi badla, lekin $I_{b} 𝐋$ ab doosri taraf point karta hai, isliye b par force ghoom kar a se door point karta hai, aur a par force bhi ulta ho jaata hai. Sirf dono currents ki relative direction matter karti hai: ek hi taraf behte currents attract karte hain; ulti taraf behte currents repel karte hain. Isi tarah lightning current ne tube ko pichkaya, kyunki uski saari strips mein current ek hi taraf beh raha tha. Dhyaan rakho ki charges wala rule yahan mat lagao, jahan signs ulta kaam karte hain.

Per unit length force, $f_{b a} = F_{b a} / L$ , hai

Parallel wires ke beech per unit length force

f_{b a} = \frac{μ_{0} I_{a} I_{b}}{2 π d}

(4.19)

Socho zaraNCERT Socho zara · p. 134

Newton's third law kab fail hota hai?

Do steady currents ki jagah do protons lo. Proton P $x$ -axis ke saath origin ki taraf chalta hai. Jis pal woh $(- d, 0)$ par pahunchta hai, doosra proton Q $+ y$ ke saath chalte hue origin se guzarta hai. Q, P ki motion ki line par theek aage hai, jahan P ka magnetic field zero hai, isliye Q par P se koi magnetic force nahi lagta. Lekin P, Q ke path ke side mein hai, isliye P par Q ka field zero nahi hai, aur P side mein dhakela jaata hai. Magnetic forces barabar aur ulte nahi hain, isliye third law akele fail ho jaata hai. Mechanics ko momentum bachata hai: electromagnetic field ka apna momentum hota hai, aur particles plus field ka total conserved rehta hai. Is chapter ke steady currents ke liye forces barabar aur ulte hi aate hain.

Ampere

1946 se 2019 tak (4.19) current ki SI unit ka aadhaar tha. Socho vacuum mein $1 m$ ki doori par rakhi do bahut lambi, bahut patli, seedhi parallel wires, jinmein barabar steady currents hain. Har ek mein current ko ek ampere (ऐम्पियर) define kiya gaya tha jab kisi bhi wire ke har metre par $2 \times 10^{- 7} N$ ka force lage. Formula se check karo: $f = (2 \times 10^{- 7}) (1) (1) / 1 = 2 \times 10^{- 7} N / m$ .

Dekho kaun si unit pehle aayi. Ampere base unit tha, aur coulomb usse banaya gaya: $1 C = 1 A s$ , yaani woh charge jo ek ampere ka steady current ek second mein kisi bhi cross-section se le jaata hai. Definition ko practice mein laana use state karne se mushkil tha, kyunki koi asli wire infinitely long nahi hoti. Metrologists ne precise coils lapeti, fixed aur movable coils mein same current bheja, aur unke beech ke pull ko current balance par taula. Forces bahut chhote hain: ek ampere ek metre ki doori par sirf $2 \times 10^{- 7} T$ ka field banata hai, Earth ke field ke ek per cent se bhi kam, isliye laboratory ke har bhatakte field ko cancel karna ya uska hisaab rakhna padta tha.

Example 4.9NCERT Example 4.9 · p. 124

Earth ke field mein DC power line

Ek high-voltage DC link overhead conductor mein $1.5 k A$ ka steady current le jaata hai. Route par Earth ke field ka horizontal part $3.5 \times 10^{- 5} T$ theek north ki taraf maano, aur uske vertical part ko ignore karo. Conductor ke har metre par magnetic force ka size aur direction nikaalo jahan line (a) theek east chalti hai, current east ki taraf, (b) theek north chalti hai, aur (c) north-east ki taraf chalti hai. (d) Conductor ke ek metre ka mass lagbhag $1.6 k g$ hai. Kya line lagane wale engineers ke liye magnetic force matter karta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Teacher demonstration

Attract aur repel karti strips

Tumhe chahiye

kitchen ki aluminium foil
ek lakdi ki rod aur do clips
ek low-voltage laboratory supply jo kuch amperes de sake
connecting leads

Pehle guess karo

Jab current ek strip mein neeche aur doosri mein upar behta hai, to unke beech ke gap ka kya hota hai?

Kam ho jaata haiBadh jaata haiNahi badalta

Aluminium foil ki lagbhag $2 c m$ chaudi aur $40 c m$ lambi do strips kaato, aur unhe lakdi ki rod se saath-saath latkao taaki unke beech lagbhag $1 c m$ ka gap ho.
Unke upar ke ends ko aapas mein jodo aur neeche ke ends ko low-voltage supply se, taaki current ek strip mein neeche aur doosri mein upar behe. Thodi der ke liye switch on karo aur gap dekho.
Ab dobara aise jodo ki current dono strips mein neeche behe (parallel mein), aur phir switch on karo.

Quick check

3 mein se 1

Do long parallel wires mein ulti directions mein currents hain. Woh kya karti hain?

Ek doosre par koi force nahi lagatiEk doosre ko attract karti hainEk doosre ko repel karti hain

Answer dekho

Answer (C) Ek doosre ko repel karti hain

Antiparallel currents repel karte hain. Parallel currents attract karte hain, jo like charges ke rule ka ulta hai.

4.9 Torque on Current Loop, Magnetic Dipole

4.9.1 Uniform magnetic field mein rectangular current loop par torque

Electric motor ki coil kyun ghoomti hai? Uniform magnetic field mein current loop ek side par ek taraf aur doosri side par ulti taraf dhakela jaata hai. Forces ke roop mein dono pushes cancel ho jaate hain, isliye poora loop kahin nahi khinchta, lekin woh use marodte hain. Uniform electric field mein electric dipole (Chapter 1) bilkul aisa hi behave karta hai: net force nahi, lekin torque. Yahi marod Section 4.10 wale meter ko chalata hai.

Current $I$ wala ek rectangular loop lo, jo apne centre se guzarne wale axis ke around ghoom sakta hai (Fig. 4.15). Uski do sides, har ek length $b$ ki, axis ke parallel hain; baaki do, har ek length $a$ ki, uske right angle par hain. Loop ka area $A = a b$ hai. Field $𝐁$ uniform hai aur axis ke perpendicular hai, aur $θ$ , $𝐁$ aur loop ke normal ke beech ka angle hai.

2/4Loop ghoomne par forces ka size aur direction same rehta hai, lekin unki lines of action paas aakar $a \sin θ$ doori par aa jaati hain: torque ghat kar $I A B \sin θ$ ho jaata hai.

Fig. 4.15Axis ke parallel do sides par forces ek couple banate hain jiska arm,

a \sin θ

𝐦

𝐁

ki taraf ghoomne par chhota hota jaata hai.NCERT Figs. 4.18, 4.19 · p. 125

Left mein ek chhota perspective sketch ek seedha khada rectangular loop dikhata hai, jiske beech se ek dashed vertical rotation axis guzarta hai, uske around current arrows hain aur upar-neeche right ki taraf point karte green field arrows B hain. Right mein loop axis ke saath upar se dikhaya gaya hai: axis ke parallel do sides ek dot, current page se bahar, aur ek cross, current page ke andar, ki tarah dikhti hain, jo pivot se guzarne wali moti line se judi hain, aur us line ke right angle par magnetic moment m ka purple arrow hai. Neeche ek green arrow B ko right point karte dikhata hai. Step 1: loop ka plane field ke saath hai, m upar point karta hai, left wale dot par orange force F1 upar aur right wale cross par F2 neeche point karta hai; label tau equals I A B. Step 2: loop clockwise ghoom gaya hai taaki m, B se 30 degrees ka angle theta banaye; forces ab bhi upar aur neeche point karte hain, aur unse guzarti dashed lines a sine theta door hain; label tau equals I A B sine theta. Step 3: m, B ke saath point karta hai, dono forces ek line par bahar kheenchte hain, tau equals 0, label stable. Step 4: m, B ke ulta point karta hai, forces ek line par andar dhakelte hain, tau equals 0, label unstable.

DerivationRectangular loop par torque

Length $a$ wali sides. In do sides mein same uniform field mein barabar currents ulti directions mein hain, isliye un par forces barabar aur ulte hain. Dono forces rotation axis ke saath, in sides ke centres se hokar lagte hain, isliye ek hi line par hain: na net force dete hain na torque.

Length $b$ wali sides. Yeh hamesha $𝐁$ ke right angle par hain, isliye har ek par lagne wale force ka size hai

F_{1} = F_{2} = I b B

$𝐁$ ke right angle par, ek-ek har direction mein. Dono forces vectors ki tarah cancel hote hain, lekin ek hi line par nahi lagte, isliye couple banate hain.

Loop ka plane field ke saath ( $θ = 90^{\circ}$ ). Har force axis se $a / 2$ doori par lagta hai, aur dono torques judte hain:

τ = F_{1} \frac{a}{2} + F_{2} \frac{a}{2} = I b B \frac{a}{2} + I b B \frac{a}{2} = I (a b) B

Loop ka plane field ke saath hone par torque

τ = I A B

(4.20)

General angle. Jab loop ka normal $𝐁$ se angle $θ$ banata hai, to forces ka size aur direction same rehta hai, lekin har line of action ab axis se sirf $(a / 2) \sin θ$ door hai:

τ = F_{1} \frac{a}{2} \sin θ + F_{2} \frac{a}{2} \sin θ = I a b B \sin θ

Current loop par torque

τ = I A B \sin θ

(4.21)

Dono results ek vector equation mein likhe ja sakte hain. Loop ka magnetic moment (चुंबकीय आघूर्ण) define karo

Current loop ka magnetic moment

𝐦 = I 𝐀

(4.22)

jahan area vector $𝐀$ ka magnitude loop ke area ke barabar hai aur woh right-hand thumb rule se chune gaye normal ke saath point karta hai. Ungliyan current ke saath modo, to thumb $𝐀$ deta hai. Tab $𝐦$ aur $𝐁$ ke beech ka angle $θ$ hai, aur

Magnetic moment par torque

𝝉 = 𝐦 \times 𝐁

(4.23)

Electric field mein electric dipole se compare karo, jis par $𝝉 = 𝐩_{e} \times 𝐄$ lagta hai: magnetic moment $𝐦$ , $𝐩_{e}$ ka role nibhata hai. (4.22) se iski SI unit ampere square metre, $A m^{2}$ , hai.

Torque do positions mein zero hai, $θ = 0$ aur $θ = 180^{\circ}$ (Fig. 4.15 ke steps 3 aur 4), lekin dono bahut alag hain. $θ = 0$ ke paas chhota sa tilt $δ$ lagbhag $m B δ$ ka torque deta hai jo loop ko wapas ghumata hai: katori ke tale mein rakhi ball ki tarah, yeh equilibrium stable hai. $θ = 180^{\circ}$ ke paas chhota sa tilt aisa torque banata hai jo tilt ko badhata hai: pahadi ki choti par tiki ball ki tarah, yeh equilibrium unstable hai. Isliye har magnetic dipole field ke saath point karna pasand karta hai, aur yahi wajah hai ki compass needle Earth ke field ke saath aakar rukti hai.

Same frame par $N$ turns lapetne se har force, aur isliye moment, $N$ se multiply ho jaata hai:

N turns wali coil ka magnetic moment

𝐦 = N I 𝐀

(4.24)

Example 4.10NCERT Example 4.10 · p. 126

Chhote satellite ko modna

Ek chhota cube-shaped satellite khud ko magnetorquer se ghumata hai: 200 turns ki flat square coil, har side $8.0 c m$ , jo uske ek face ke andar fixed hai. Uske orbit mein Earth ka field lagbhag $3.0 \times 10^{- 5} T$ hai, aur jis axis par satellite ghoomta hai uske around uska moment of inertia $2.2 \times 10^{- 3} k g m^{2}$ hai. Coil mein $0.16 A$ current hai. (a) Coil ka magnetic moment nikaalo. (b) Jab coil ka axis field se $60^{\circ}$ par ho, to satellite par torque nikaalo. (c) $60^{\circ}$ par rest se shuru karke, koi aur torque na ho to, jab coil ka axis field ke saath line up hota hai tab satellite kitni tezi se ghoom raha hota hai? (d) Dikhao ki chhote angles ke liye satellite field direction ke around aage-peeche jhoolta hai, aur is jhoolne ka period nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4.11NCERT Example 4.11 · p. 127

Jhukna, ghoomna nahi

40 turns ki flat rectangular coil, $5.0 c m \times 4.0 c m$ , mein $0.50 A$ current hai. Yeh level, frictionless table par rakhi hai aur upar se dekhne par current anticlockwise hai. Axes lo: $x$ east, $y$ north aur $z$ upar. Uniform field $𝐁 = (0.30 \hat{𝐢} + 0.40 \hat{𝐤}) T$ on kiya jaata hai. (a) Coil ka magnetic moment aur us par torque nikaalo. Kya koi bhi uniform field, kisi bhi direction mein, coil ko table par turntable ki tarah ghuma sakta hai? (b) Agar coil kisi bhi direction mein ghoom sake, to woh kahan aakar rukegi, aur uske centre par uska apna field wahan ke $𝐁$ se kaise compare karega?

(c) Coil ki jagah $18 c m$ lambi (rectangle ka perimeter) soft, flexible wire ka single turn rakha jaata hai, jo table par dheela pada hai aur usi sense mein $0.50 A$ le ja raha hai, taaki uska moment upar point kare. Wire kaun si shape leti hai, aur usi wire ko $5.0 c m \times 4.0 c m$ rectangle mein modne ke muqable uska magnetic moment kitne factor se badalta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Torque north point karta hai, jo horizontal hai. Yeh coil ko north–south line ke around ghumata hai, uska west edge uthata hai taaki $𝐦$ east ki taraf, $𝐁$ ki taraf, jhuke. Coil ko turntable ki tarah ghumane ke liye vertical torque chahiye. Lekin $𝐦 \times 𝐁$ hamesha $𝐦$ ke perpendicular hai, jo vertical hai, isliye $𝐁$ ki direction kuch bhi ho, torque horizontal hai: koi uniform field yeh nahi kar sakta.

Quick check

3 mein se 1

20 turns ki flat coil, har turn ka area

4.0 \times 10^{- 3} m^{2}

0.30 T

ke field mein

0.50 A

le jaati hai. Us par sabse bada torque kitna lag sakta hai, newton metres mein?

N m

Answer dekho

Answer 0.012 N m

τ_{\max} = N I A B = (20) (0.50) (4.0 \times 10^{- 3}) (0.30) = 1.2 \times 10^{- 2} N m

4.9.2 Magnetic dipole ke roop mein circular current loop

Chhote current loop se kaafi door hat jao, to uska field electric dipole ke field se bilkul alag nahi pehchana ja sakta. Axial field, (4.11), lo. Bahut door $x ≫ R$ , isliye $x^{2} + R^{2} \approx x^{2}$ , aur loop ke area ke liye $A = π R^{2}$ lekar:

B \approx \frac{μ_{0} I R^{2}}{2 x^{3}} = \frac{μ_{0} I A}{2 π x^{3}}

$m = I A$ ke saath axis par yeh ban jaata hai

Loop ke axis par door ka field

𝐁 \approx \frac{μ_{0}}{4 π} \frac{2 𝐦}{x^{3}} (x ≫ R)

(4.25(a))

Ab short electric dipole ke axis par field yaad karo, $𝐄 = \frac{1}{4 π ε_{0}} \frac{2 𝐩_{e}}{x^{3}}$ (Chapter 1). Ek chhoti dictionary se translate karo to dono formulas same hain: $𝐄 \to 𝐁$ , $𝐩_{e} \to 𝐦$ aur $1 / ε_{0} \to μ_{0}$ . Yeh dictionary doosri directions mein bhi kaam karti hai. Electric dipole ke perpendicular bisector par field, $- 𝐩_{e} / (4 π ε_{0} x^{3})$ , ko translate karne par loop ke apne plane mein door ke point par uska field milta hai:

Loop ke plane mein door ka field

𝐁 \approx - \frac{μ_{0}}{4 π} \frac{𝐦}{x^{3}} (x ≫ R)

(4.25(b))

Loop ko chhota karte jao aur current badhate jao taaki $m = I A$ same rahe, to approximations exact ho jaate hain: limit ek point magnetic dipole hai. Loop ka circular hona bhi zaroori nahi. Koi bhi flat current loop, door se dekhne par, apne moment $𝐦 = I 𝐀$ ka dipole field banata hai, aur yeh moment $𝐩_{e}$ ka magnetic partner hai.

2/2Paas se: charges ke beech electric field (orange arrowhead) $+$ se $-$ ki taraf, $𝐩_{e}$ ke ulta, point karta hai; loop ke andar magnetic field (orange arrowhead) $𝐦$ ke saath point karta hai.

Fig. 4.16Door se current loop aur electric dipole ke field patterns bilkul ek jaise hain. Paas se woh alag hain: electric field lines charges par khatam hoti hain, lekin magnetic field lines loop se guzar kar band ho jaati hain.NCERT p. 128

Do square panels saath-saath, label electric dipole aur current loop. Step 1, door se: dono panels mein blue field lines ka same pattern hai, aise loops jo centre se axis ke saath right ki taraf nikalte hain aur mud kar left se wapas ghuste hain, un par arrows hain aur centre mein source ke liye chhota dot hai. Step 2, paas se: electric dipole panel mein right par positive charge aur left par negative charge hai, field lines plus charge se shuru hokar minus charge par khatam hoti hain, aur charges ke beech ek orange arrow left point karta hai. Current loop panel mein upar dot aur neeche cross hai, yaani loop ke kate ends, field lines har end ke around ghoomti hain aur loop se seedhi guzarti hain, aur unke beech ek orange arrow right point karta hai.

Source ke paas yeh samaanta toot jaati hai (Fig. 4.16). Electric dipole do alag charges ki jodi hai, aur uski field lines ek par shuru hokar doosre par khatam hoti hain; charges ke beech field $𝐩_{e}$ ke ulta point karta hai. Current loop ke alag parts nahi hain: uski field lines loop se seedhi, $𝐦$ ke saath, guzarti hain aur apne aap mein band ho jaati hain. Akele charge ka magnetic version, magnetic monopole, kabhi nahi mila. Isliye jahan electric dipole do simple pieces se banta hai, wahin magnetic dipole khud hi sabse simple known magnetic object hai.

Ampère is analogy ko poori had tak le gaye. Current loop chhote magnet jaisa field banata hai (Fig. 4.9) aur field mein compass needle ki tarah ghoomta hai, to shayad har magnet asal mein chhote-chhote ghoomte currents ka collection hai. Yeh idea aadha sahi hai. Atoms ke around ghoomte electrons sach mein chhote current loops ki tarah behave karte hain. Lekin electron ka apna ek magnetic moment bhi hota hai, jo uske charge jitna hi uska hissa hai, aur ghoomte charge ki koi classical picture use samjha nahi sakti. Chapter 5 yeh kahani aage badhata hai.

Example

Dipole approximation kitna achha hai?

Ek chhoti flat coil mein radius $1.0 c m$ ke 100 turns hain aur usmein $0.50 A$ current hai. Uska magnetic moment nikaalo, aur centre se $20 c m$ door uske axis par exact field ko dipole approximation se compare karo. Coil ke plane mein same doori par field kitna hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Chhote current loop ke axis par bahut door magnetic field doori

x

par kaise depend karta hai?

1 / x

ki tarah

1 / x^{2}

ki tarah

1 / x^{3}

ki tarah

Answer dekho

Answer (C) $1 / x^{3}$ ki tarah

(4.25(a)) dikhata hai

B \propto 1 / x^{3}

, bilkul electric dipole ke field ki tarah.

1 / x^{2}

single point charge ka field hai, jiska koi magnetic counterpart nahi hai.

4.10 The Moving Coil Galvanometer

Needle wala (analogue) multimeter torch circuit mein lagao, to woh $0.35 A$ padh sakta hai; use voltage range par switch karo to bulb ke across $2.9 V$ dikha sakta hai. Aise meter ke andar, aur har classic laboratory ammeter aur voltmeter mein, ek hi device hai: moving coil galvanometer (चल कुंडली धारामापी). Yeh current ko marod mein badalta hai, bilkul usi coil par torque se jo humne abhi nikaala.

Iska dil ek halki rectangular coil hai jismein patli insulated wire ke kai turns hain, aur jo pivot par aise lagi hai ki ek fixed axis ke around ghoom sake (Fig. 4.17). Coil permanent magnet ke curved pole pieces aur coil ke andar soft iron ke fixed cylinder ke beech ke patle gap mein baithi hai. Is arrangement se gap mein field strong aur radial hota hai: coil ki sides jahan bhi hon, wahan field lines cylinder ke radius ke saath point karti hain. Ek patli spiral hairspring coil ke ghoomne ko rokti hai, aur coil se juda ek halka pointer scale par chalta hai.

4/4Coil tab tak ghoomti hai jab tak spring ka restoring torque $k ϕ$ , $N I A B$ ke barabar na ho jaaye; pointer deflection $ϕ$ dikhata hai.

Fig. 4.17Radial field mein coil par har angle par same torque

N I A B

lagta hai, aur woh tab tak ghoomti hai jab tak spring ka torque

k ϕ

use balance na kar de.NCERT Fig. 4.20 · p. 130

Instrument ka top view. Left mein N label wala block aur right mein S label wala block, jinke curved faces beech mein grey circular soft-iron core ke around hain; coil core se guzarti ek patli bar hai jiski do sides curved gaps mein baithi hain, aur centre mein ek chhota spiral spring aur pivot hai. Pivot se ek pointer neeche right ki taraf jaata hai, neeche ke curved scale ke upar. Step 1: parts par labels hain coil, core, spring, scale aur pointer. Step 2: dono gaps mein chhote blue arrows radii ke saath point karte hain, N pole se core mein aur core se bahar S pole tak, jo radial field dikhate hain. Step 3: current behta hai, left coil side par dot aur right par cross se dikhaya gaya; orange arrows left side par upar aur right side par neeche force dikhate hain, jo coil ko clockwise ghumata hai, label tau equals N I A B. Step 4: coil aur pointer angle phi se clockwise ghoom gaye hain, jo pointer ki dashed rest position se ek arc se mark hai, aur label hai k phi equals N I A B.

DerivationGalvanometer ka deflection

Kyunki field radial hai, coil kitne bhi angle se ghoom jaaye, uske plane mein hamesha field lines hoti hain. Isliye (4.21) mein $\sin θ = 1$ , aur $N$ turns wali coil, har turn ka area $A$ , par magnetic torque lagta hai

τ = N I A B

Spring marodti hai aur marod ke proportional torque, $k ϕ$ , se wapas dhakelti hai, jahan $ϕ$ deflection ka angle hai. Constant $k$ spring ki stiffness naapta hai: yeh use ek radian marodne ke liye zaroori torque hai, aur ise torsional constant kehte hain. Coil us angle par rukti hai jahan dono torques balance hote hain:

k ϕ = N I A B

Moving coil galvanometer ka deflection

ϕ = (\frac{N A B}{k}) I

(4.26)

Bracket ki har cheez instrument bante samay fix ho jaati hai, isliye $ϕ \propto I$ aur scale ko barabar divisions mein mark kiya ja sakta hai. Linear scale hi field ko radial banane ki main wajah hai.

Kabhi humein koi number nahi chahiye, bas haan ya na: kya koi current beh raha hai, aur kis taraf? Wheatstone bridge (Chapter 3) ke balance point par jawab hona chahiye: koi nahi. Is tarah current detector ke roop mein use hone wale galvanometer ka zero scale ke beech mein hota hai, taaki left ya right swing turant bata de ki bridge balance ki kis side par hai.

Ammeter ke roop mein galvanometer

Maan lo tum lagbhag $1 A$ current aise galvanometer se naapna chahte ho jo $2 m A$ par full scale pahunchta hai aur jiski coil resistance $R_{G} = 25 Ω$ hai. Turant do problems aati hain. Current 500 guna zyada hai: pointer apne stop se zor se takrayega aur patli coil jal sakti hai. Aur ammeter circuit ke saath series mein lagta hai, isliye uske $25 Ω$ circuit ke resistance mein judkar wahi current kam kar denge jo tum naapna chahte ho.

Ek chhota resistor dono problems hal karta hai. Shunt (शंट), yaani coil ke parallel mein laga ek low resistance $r_{s}$ , lagbhag saara current le leta hai, aur sirf ek chhota, fixed fraction coil se guzarta hai (Fig. 4.18). Tab meter ka total resistance hai

\frac{R_{G} r_{s}}{R_{G} + r_{s}} \approx r_{s} (R_{G} ≫ r_{s})

Hamare $2 m A$ , $25 Ω$ galvanometer ke liye $1 A$ range par numbers daalo. Shunt lagbhag $0.05 Ω$ aata hai (tareeka neeche wale example mein hai), isliye tayyar ammeter circuit mein sirf $0.05 Ω$ jodta hai: $5 Ω$ ya usse zyada ke circuit mein yeh current ko 1% ya usse kam ghatata hai. Coil hamesha current ka same hissa leti hai, yahan $r_{s} / (R_{G} + r_{s}) = 1 / 500$ , isliye pointer ki position total current batati hai aur scale amperes mein mark kiya ja sakta hai.

2/3Parallel mein chhota shunt $r_{s}$ current $I$ ka zyadatar hissa le jaata hai; sirf ek chhota, fixed fraction coil se behta hai.

Fig. 4.18Parallel mein chhota shunt galvanometer ko ammeter banata hai; series mein bada resistance use voltmeter banata hai.NCERT Figs. 4.21, 4.22 · p. 130

Do terminals ke beech ek seedhi circuit line hai, jis par G likha circle aur uske paas chhota zigzag resistor R G hai. Step 1: akela galvanometer, chhote current I G ke liye patle orange arrow ke saath jo full-scale deflection deta hai. Step 2: G ke neeche parallel mein ek green zigzag resistor r s joda gaya hai, aur dono ke around dashed box par ammeter likha hai; ek mota arrow I andar aata hai, ek patla arrow I G, G se guzarta hai aur ek mota arrow I minus I G shunt se guzarta hai. Step 3: shunt hat gaya hai aur G ke baad series mein ek lamba purple zigzag resistor R joda gaya hai, dono ke around dashed box par voltmeter likha hai; sirf chhota current I G behta hai.

Definition

Current sensitivity

Coil se guzarne wale har unit current ke liye pointer kitna ghoomta hai: $ϕ / I = N A B / k$ .

ExampleJo galvanometer $100 μ A$ ke liye $0.5 r a d$ deflect hota hai, uski current sensitivity $5000 r a d / A$ hai.

Current sensitivity

\frac{ϕ}{I} = \frac{N A B}{k}

(4.27)

(4.27) ko design guide ki tarah padho: strong magnet, badi coil, naram hairspring ya zyada turns, har ek per ampere deflection badhata hai. Har ek ki keemat hai. Naram spring se pointer rukne se pehle zyada der hilta hai aur meter aasaani se overload hota hai, aur zyada turns resistance badhate hain, jo tab matter karta hai jab galvanometer voltmeter banta hai, jaisa hum aage dekhenge.

SimulationShunt ke saath galvanometer

Ek circuit jismein 4.5 volt cell, 5 ohm load resistor, aur 0.2 ohm shunt ke saath parallel mein judi 40 ohm galvanometer coil hai, aur uske baad ek ideal ammeter jo total current padhta hai. Har resistor apne across voltage aur apne through current dikhata hai. Tum kisi resistor ko tap karke select kar sakte ho aur slider se uska resistance badal sakte ho, cell voltage badal sakte ho, switch khol aur band kar sakte ho, aur moving charges dikha sakte ho.

Fig. 4.19Ek

4.5 V

cell

5 Ω

load aur

0.2 Ω

shunt wali galvanometer coil (

40 Ω

) se current chalata hai. Kisi resistor ko tap karke uska current padho aur uski value badlo.

Try karo

G coil aur shunt se guzarta current padho. Total ka kitna fraction coil se jaata hai? Ise $r_{s} / (R_{G} + r_{s})$ se compare karo.
Shunt select karo aur use sabse badi value tak badhao, jo use hata dene jaisa hai. Ab ammeter kya padhta hai? Bina kisi meter ke current $0.90 A$ hota.
Reset button dabakar shunt ko wapas $0.2 Ω$ par lao, phir load select karke use lagbhag $50 Ω$ tak badhao. Bina meter ke ab current $0.090 A$ hota. Kya current par meter ka effect pehle se zyada hai ya kam?

Voltmeter ke roop mein galvanometer

Ab maan lo tumhe circuit mein resistor $R_{1}$ ke across voltage chahiye. Voltmeter resistor ke across, parallel mein, lagaya jaata hai, aur jo bhi current woh leta hai, woh resistor se chura leta hai, jisse wahi voltage kam ho jaata hai jo tum padhna chahte ho. Resistance $R_{V}$ wala meter lagao to dono milkar single resistance $R_{1} R_{V} / (R_{1} + R_{V})$ ki tarah behave karte hain; $R_{V} = 100 R_{1}$ ke saath yeh $100 R_{1} / 101$ hai, $R_{1}$ se lagbhag 1% kam. Isliye achhe voltmeter ka resistance us cheez se bahut zyada hona chahiye jiske across woh laga hai. Yeh hum coil ke saath series mein bada resistance $R$ jodkar paate hain, jisse meter ka total resistance $R_{G} + R \approx R$ ho jaata hai.

Definition

Voltage sensitivity

Meter ke across har volt ke liye pointer kitna ghoomta hai: $ϕ / V = N A B / k R$ , jahan $R$ meter ka total resistance hai.

Voltage sensitivity

\frac{ϕ}{V} = (\frac{N A B}{k}) \frac{I}{V} = (\frac{N A B}{k}) \frac{1}{R}

(4.28)

Kya current ke liye zyada sensitive galvanometer voltage ke liye bhi zyada sensitive hota hai? Zaroori nahi. Table 4.4 ek coil ko uske us version se compare karta hai jise same frame par same wire ke dugne turns se dobara lapeta gaya hai.

Table 4.4Galvanometer coil ke turns dugne karna

Coil	Turns	Resistance	Current sensitivity	Voltage sensitivity
Original	$N$	$R$	$N A B / k$	$N A B / k R$
Dobara lapeti	$2 N$	lagbhag $2 R$ (dugni wire)	$2 N A B / k$ (dugna)	$2 N A B / k (2 R)$ (same)

$N$ dugna karne se per ampere torque dugna hota hai, lekin wire ki length bhi dugni ho jaati hai, isliye resistance bhi dugna hota hai, aur diya hua voltage coil se jitna current chala sakta hai woh aadha ho jaata hai. To jo cheez galvanometer ko ammeter ke roop mein behtar banati hai, woh use apne aap voltmeter ke roop mein behtar nahi banati.

Compare

Ek galvanometer se ammeter aur voltmeter

Ammeter

Circuit ke saath series mein juda
Coil ke parallel mein chhota shunt $r_{s}$
Total resistance bahut chhota, lagbhag $r_{s}$
Ideal ammeter ka resistance zero

Voltmeter

Component ke saath parallel mein juda
Coil ke saath series mein bada resistance $R$
Total resistance bahut bada, lagbhag $R$
Ideal voltmeter ka resistance infinite

Example 4.12NCERT Example 4.12 · p. 131

Kaun si range zyada sahi reading deti hai?

$3.0 V$ emf aur $0.50 Ω$ internal resistance wala torch cell $5.5 Ω$ resistance ka bulb jalata hai. Ek student multimeter se current naapta hai, pehle uski $2 A$ range par, jahan meter ka resistance (galvanometer plus shunt) $0.20 Ω$ hai, aur phir uski $10 A$ range par, jahan yeh $0.020 Ω$ hai. (a) Circuit mein bina meter ke current kitna hai? (b) Har range par meter kya padhta hai, aur har reading kitne percent kam hai? (c) Error 1% se kam rakhne ke liye meter ka maximum resistance kitna ho sakta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Ek galvanometer, do meters

Ek galvanometer ki coil resistance $25 Ω$ hai aur yeh $2.0 m A$ current par full-scale deflection deta hai. Ise (a) $5.0 A$ tak padhne wala ammeter aur (b) $10 V$ tak padhne wala voltmeter banane ke liye isse kya, aur kaise, jodna chahiye?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Galvanometer kitna sensitive hai?

Ek galvanometer coil mein 200 turns hain, har turn ka area $2.0 c m^{2}$ , aur yeh $0.25 T$ ke radial field mein hai. Uski hairspring ka torsional constant $2.0 \times 10^{- 6} N m / r a d$ hai. Iski current sensitivity, ise $30^{\circ}$ ghumane wala current, aur agar scale $90^{\circ}$ cover karta hai to iska full-scale current nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Moving coil galvanometer ke pole faces curved kyun hote hain, aur coil ke andar soft-iron cylinder kyun hota hai?

Field ko radial banane ke liye, taaki torque coil ke angle par depend na kareCoil ko zyada ghoomne se rokne ke liyeCoil ko halka banane ke liye

Answer dekho

Answer (A) Field ko radial banane ke liye, taaki torque coil ke angle par depend na kare

Radial field mein coil ka plane hamesha field ke saath rehta hai, isliye har angle par

τ = N I A B

aur

ϕ \propto I

. Iron field ko strong bhi karta hai.

Table 4.5Is chapter ki quantities, typical sizes ke saathNCERT p. 134

Quantity	Symbol	SI unit	Base units mein	Ek typical value
Magnetic field (vector)	$𝐁$	$T$ (tesla)	$k g s^{- 2} A^{- 1}$	Earth ki surface: lagbhag $5 \times 10^{- 5} T$
Permeability of free space	$μ_{0}$	$T m A^{- 1}$	$k g m s^{- 2} A^{- 2}$	$4 π \times 10^{- 7} T m A^{- 1}$
Magnetic moment (vector)	$𝐦$	$A m^{2}$ , $J / T$ ke barabar	$A m^{2}$	Example 4.10 ki satellite coil: $0.2 A m^{2}$
Cyclotron frequency	$ν_{c}$	$H z$	$s^{- 1}$	$1 T$ mein proton: $15 M H z$
Torsional constant	$k$	$N m r a d^{- 1}$	$k g m^{2} s^{- 2}$	galvanometer hairspring: lagbhag $10^{- 6} N m / r a d$

Flash cards · 12

Jaldi yaad karo

1 / 12

Recap

Summary aur key points

Current apne aas-paas magnetic field banata hai, jaisa 1820 mein Oersted ke compass ne dikhaya (Section 4.1).
Charge par Lorentz force $𝐅 = q (𝐄 + 𝐯 \times 𝐁)$ hai. Magnetic part $𝐯$ ke perpendicular hai aur koi work nahi karta; iska size $q v B \sin θ$ hai, aur $B$ tesla mein naapa jaata hai (Section 4.2.2).
Uniform field mein current $I$ wale seedhe conductor par $𝐅 = I 𝐥 \times 𝐁$ lagta hai (Section 4.2.3).
Uniform field mein charge radius $m v / q B$ ke circle mein cyclotron frequency $ν_{c} = q B / 2 π m$ par chalta hai, jo speed par depend nahi karti; $𝐁$ ke saath velocity component ho to path pitch $2 π m v_{∥} / q B$ wala helix hai (Section 4.3).
Biot–Savart law current element ka field deta hai: $d 𝐁 = (μ_{0} / 4 π) I d 𝐥 \times 𝐫 / r^{3}$ , jahan $μ_{0} = 4 π \times 10^{- 7} T m / A$ (Section 4.4).
Radius $R$ ke loop ke axis par field $μ_{0} I R^{2} / 2 (x^{2} + R^{2})^{3 / 2}$ hai, jo centre par $μ_{0} I / 2 R$ hai (Section 4.5).
Ampère's circuital law, $\oint 𝐁 \cdot d 𝐥 = μ_{0} I$ , achhe se chune loop ke liye $B L = μ_{0} I_{e}$ ban jaata hai, aur long straight wire ke liye $B = μ_{0} I / 2 π r$ deta hai (Section 4.6).
Per unit length $n$ turns wale long solenoid ke andar field uniform hai, $B = μ_{0} n I$ , aur bahar lagbhag zero hai (Section 4.7).
Parallel currents attract karte hain aur antiparallel currents repel karte hain, per unit length force $μ_{0} I_{a} I_{b} / 2 π d$ ke saath (Section 4.8).
Current loop ka magnetic moment $𝐦 = N I 𝐀$ hai. Uniform field mein us par net force nahi lagta lekin torque $𝝉 = 𝐦 \times 𝐁$ lagta hai, aur $𝐦$ , $𝐁$ ke saath ho to woh stable equilibrium mein hai (Section 4.9).
Door se current loop ek magnetic dipole hai jiska field $𝐩_{e} \to 𝐦$ aur $1 / ε_{0} \to μ_{0}$ ke saath electric dipole ke field se match karta hai (Section 4.9.2).
Moving coil galvanometer mein torque $N I A B$ ko spring balance karti hai, $k ϕ = N I A B$ ; parallel mein chhota shunt ise ammeter aur bada series resistance voltmeter banata hai (Section 4.10).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 4.1	Practice Q1
NCERT Exercises 4.2	Practice Q2
NCERT Exercises 4.3	Practice Q3
NCERT Exercises 4.4	Practice Q4
NCERT Exercises 4.5	Practice Q5
NCERT Exercises 4.6	Practice Q6
NCERT Exercises 4.7	Practice Q7
NCERT Exercises 4.8	Practice Q8
NCERT Exercises 4.9	Practice Q9
NCERT Exercises 4.10	Practice Q10
NCERT Exercises 4.11	Practice Q11
NCERT Exercises 4.12	Practice Q12
NCERT Exercises 4.13	Practice Q13

Quick check: MCQs

4.1 Introduction3 questions
4.2.2 Magnetic field, Lorentz force4 questions
4.2.3 Current-carrying conductor par magnetic force3 questions
Angle par velocity: helix4 questions
4.4 Magnetic Field due to a Current Element, Biot–Savart Law3 questions
4.5 Magnetic Field on the Axis of a Circular Current Loop3 questions
Long straight wire ka field4 questions
4.7 The Solenoid3 questions
Ampere3 questions
4.9.1 Uniform magnetic field mein rectangular current loop par torque3 questions
4.9.2 Magnetic dipole ke roop mein circular current loop3 questions
Voltmeter ke roop mein galvanometer4 questions

Glossary · 15

Key terms (English + Hindi)

ampereऐम्पियरaimpiyar: Current ki SI unit, jo pehle $1 m$ door do parallel wires ke beech per metre $2 \times 10^{- 7} N$ ke force se define hoti thi.Chapter mein pehli baar yahan aaya →
Ampère's circuital lawऐम्पियर का परिपथीय नियमaimpiyar ka paripathiya niyam: Kisi bhi closed curve ke around $𝐁$ ka line integral $μ_{0}$ times usse guzarne wale net current ke barabar hai: $\oint 𝐁 \cdot d 𝐥 = μ_{0} I$ .
Amperian loopऐम्पियरी पाशaimpiyari paash: Aisa chuna gaya closed curve jiske har part par $𝐁$ tangential aur constant, perpendicular, ya zero ho, jisse Ampère's law lagana aasaan ho jaata hai.Chapter mein pehli baar yahan aaya →

Biot–Savart lawबायो-सावर्ट नियमbayo-savart niyam: Current element ka field: $d 𝐁 = (μ_{0} / 4 π) I d 𝐥 \times 𝐫 / r^{3}$ .

current sensitivityधारा सुग्राहिताdhaara sugraahita: Galvanometer ka per unit current deflection, $ϕ / I = N A B / k$ .
cyclotron frequencyसाइक्लोट्रॉन आवृत्तिcyclotron aavritti: Woh frequency $ν_{c} = q B / 2 π m$ jisse charged particle uniform magnetic field mein ghoomta hai; yeh particle ki speed par depend nahi karti.Chapter mein pehli baar yahan aaya →

Lorentz forceलोरेंज बलlorenj bal: Dono fields se guzarte point charge par lagne wala combined electric aur magnetic force $q (𝐄 + 𝐯 \times 𝐁)$ .

magnetic momentचुंबकीय आघूर्णchumbakiya aaghurn: Flat coil ke liye $𝐦 = N I 𝐀$ , $A m^{2}$ mein; field $𝐁$ mein is par torque $𝐦 \times 𝐁$ lagta hai.Chapter mein pehli baar yahan aaya →
moving coil galvanometerचल कुंडली धारामापीchal kundali dhaaraamaapi: Aisa instrument jismein radial magnetic field mein rakhi coil spring ke khilaaf current ke proportional angle se ghoomti hai.Chapter mein pehli baar yahan aaya →

permeability of free spaceनिर्वात की चुंबकशीलताnirvaat ki chumbakasheelta: Magnetism ke laws mein constant $μ_{0} = 4 π \times 10^{- 7} T m / A$ , jo wahi role nibhata hai jo electrostatics mein $ε_{0}$ .Chapter mein pehli baar yahan aaya →
pitch (helix ki)चूड़ी अंतरालpitch (of a helix)choodi antaral: Woh doori jo charged particle apne helical path ke ek turn mein magnetic field ke saath tay karta hai, $p = v_{∥} T$ .Chapter mein pehli baar yahan aaya →

shuntशंटshunt: Galvanometer coil ke across, parallel mein, laga low resistance jo zyadatar current le jaata hai aur galvanometer ko ammeter bana deta hai.Chapter mein pehli baar yahan aaya →
solenoidपरिनालिकाparinaalika: Paas-paas turns wale helix mein lapeti lambi wire; long solenoid ke andar field uniform hai, $B = μ_{0} n I$ .Chapter mein pehli baar yahan aaya →

teslaटेस्लाtesla: Magnetic field ki SI unit: $1 T = 1 N A^{- 1} m^{- 1}$ . Ek gauss $10^{- 4} T$ hai.Chapter mein pehli baar yahan aaya →

voltage sensitivityवोल्टता सुग्राहिताvoltata sugraahita: Galvanometer ke across per unit voltage uska deflection, $ϕ / V = N A B / k R$ .

Moving Charges and Magnetism: Class 12 Physics Chapter 4 Hinglish mein

Is chapter mein aap kya seekhenge

4.1 Introduction

4.2 Magnetic Force

4.2.1 Sources aur fields

4.2.2 Magnetic field, Lorentz force

4.2.3 Current-carrying conductor par magnetic force

4.3 Motion in a Magnetic Field

Field ke right angle par velocity

Angle par velocity: helix

4.4 Magnetic Field due to a Current Element, Biot–Savart Law

4.5 Magnetic Field on the Axis of a Circular Current Loop

4.6 Ampère's Circuital Law

Long straight wire ka field

4.7 The Solenoid

4.8 Force between Two Parallel Currents, the Ampere

Ampere

4.9 Torque on Current Loop, Magnetic Dipole

4.9.1 Uniform magnetic field mein rectangular current loop par torque

4.9.2 Magnetic dipole ke roop mein circular current loop

4.10 The Moving Coil Galvanometer

Ammeter ke roop mein galvanometer

Voltmeter ke roop mein galvanometer

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye