Moving Charges and Magnetism: Class 12 Physics · Chapter 4
Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗
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What you will learn in Moving Charges and Magnetism
- Use the Lorentz force and F = I l × B to find the size and direction of magnetic forces on moving charges and current-carrying wires
- Explain why a charge in a uniform magnetic field moves in a circle or a helix, and calculate its radius, period, cyclotron frequency and pitch
- State the Biot–Savart law and use it to find the field at any point on the axis of a current-carrying ring
- Apply Ampère's circuital law to find the fields of a long straight wire, a thick wire and a long solenoid
- Derive the force between two parallel currents and relate it to the definition of the ampere
- Derive the torque on a current loop, define its magnetic moment and explain why a small loop behaves as a magnetic dipole
- Explain how a moving coil galvanometer works and how it is converted into an ammeter or a voltmeter
Original companion text aligned to NCERT Physics Class 12, Chapter 4, pp. 107–135. Not affiliated with or endorsed by NCERT.
4.1 Introduction
Every electric motor, loudspeaker, hospital MRI scanner and particle accelerator depends on a link that stayed hidden until 1820: an electric current produces a magnetic field. Before then the two subjects had separate histories and separate toolkits. Electricity meant rubbed amber, sparks and Leyden jars; magnetism meant lodestones and the compass needles that sailors steered by.
Hans Christian Oersted, a professor in Copenhagen, came across it in 1820 while showing an experiment to his students, and it is easy to see for yourself (Fig. 4.1). Lay a wire across the top of a compass, parallel to its needle, and connect it for a moment to a cell. The needle kicks sideways and stays turned for as long as the current flows. Swap the connections and it kicks the other way. The kick grows with the current, and shrinks as the wire is lifted away from the needle.
2/4A current flowing north swings the north end of the needle towards the west.
A top view of a compass with W and E marked, and a straight copper-coloured wire running from north to south across it, just above the needle. A north arrow is drawn beside it. Step 1: no current flows and the needle, solid at its north end and hollow at its south end, points north along the wire. Step 2: arrows on the wire show a current flowing north, and the north end of the needle has swung about 50 degrees towards the west. Step 3: the arrows point south, and the needle has swung the same amount towards the east. Step 4: double arrows show twice the current flowing north, and the needle swings further, about 70 degrees towards the west.
Where exactly is the needle trying to point? Stand a current-carrying wire upright through a card, and surround it with a ring of small compasses or a sprinkling of iron filings. The needles and the filings line up along circles centred on the wire. So the field of a straight current does not point away from its source, the way the electric field of a charge does; it wraps round it. A current, then, does more than warm a wire or light a bulb: it fills the space around it with a magnetic field. That discovery is the starting point of this chapter.
From a swinging needle to electromagnetism
- 1820
Oersted's needle
A current in a wire turns a nearby compass needle, the first evidence that electricity and magnetism are linked.
- 1820
Biot and Savart
Measurements with magnetised needles give the law for the field of a small piece of current.
- 1820–27
Ampère's experiments
Parallel currents attract, antiparallel currents repel; Ampère builds a mathematical theory of forces between currents.
- 1864
Maxwell's equations
Electricity, magnetism and light are brought together in one theory of the electromagnetic field.
- 1887–95
Radio waves
Heinrich Hertz makes and detects radio waves in Germany; in Kolkata, J. C. Bose generates and studies millimetre-wavelength waves, and Marconi begins sending radio signals.
- 1895
Lorentz force
Lorentz writes the force on a moving charge in electric and magnetic fields in the form used today.
- 1931
The cyclotron
Ernest Lawrence uses the circling of charged particles in a magnetic field to accelerate them to high energies.
Two questions run through the chapter. How does a magnetic field push on a moving charge, or on a wire carrying a current? And how does a current create a magnetic field in the first place? The answers explain why charged particles circle and spiral in a field (the idea behind the cyclotron), why neighbouring wires pull on each other, and how a small coil turning in a magnet becomes the moving coil galvanometer used to measure currents and voltages.
Many diagrams in this chapter show currents and fields that point straight into or out of the page, and we draw them with two symbols borrowed from archery (Fig. 4.2). A circle with a dot marks something coming out of the page, towards you, like the point of an arrow flying straight at you. A circle with a cross marks something going into the page, away from you, like the crossed flights at the tail of an arrow that is flying away.
On the left, a circle with a dot at its centre, labelled out of the page, towards you. On the right, a circle with a diagonal cross, labelled into the page, away from you. Between them an arrow is drawn sideways with its point at the left, labelled tip, and its crossed tail feathers at the right, labelled tail, linking each end of the arrow to the symbol it looks like head-on.
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Answer (B) It swings the same amount the other way
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Answer (A) A circle with a dot at its centre
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Answer False
4.2 Magnetic Force
4.2.1 Sources and fields
In Chapter 1 we split the electric interaction into two steps. A charge first fills the space around it with an electric field; a second charge then responds to whatever field it finds at its own position. For a point charge at the origin, the field at position is
- the unit vector pointing from towards the point
and the force on a charge placed there is
Chapter 1 left us two ideas worth keeping. The first is superposition: the field of many sources is the vector sum of the fields they would make one at a time. The second is that a field is more than bookkeeping. When charges jiggle in a mobile-phone tower, the charges in your phone respond only after the disturbance has crossed the gap at the speed of light, carrying energy and momentum with it. Magnetic fields share both properties. Here, though, every current is steady, so every field stays the same from moment to moment and no delays arise.
Magnetism works in the same two steps. Moving charges, and so currents, set up a magnetic field (चुंबकीय क्षेत्र), a vector with a size and direction at every point in space; any other moving charge then responds to the field it finds at its own position. What is new is how it responds, and that is where we begin.
4.2.2 Magnetic field, Lorentz force
Fire a charged particle through a region that contains both an electric field and a magnetic field . The electric force is the familiar . The magnetic force turns out to depend on the particle's velocity as well, through a cross product. The complete force on a point charge at position is
The total force that electric and magnetic fields exert together on a point charge travelling at velocity .
ExampleA proton moving at at right angles to a field feels a magnetic force of .
This expression, named after the Dutch physicist Hendrik Lorentz, summarises many experiments by Ampère and his successors. Its magnetic part, , is strikingly unlike the electric part:
- A charge at rest feels no magnetic force at all, however strong the field. The force exists only while the charge moves.
- The force is always at right angles to the velocity and to the field, in the direction of for a positive charge (Fig. 4.3).
- Its size, , depends on the angle between and . Motion along the field lines, forwards or backwards, gives no force.
- Changing the sign of the charge reverses the force: an electron and a proton moving side by side are pushed in opposite directions.
1/4With at right angles to the force is largest, . It is perpendicular to the plane of and : turning from (into the picture) to (to the right) sends the right-hand thumb downwards.
On the left a flat shaded plane is drawn in perspective with a charge on it, a green arrow for the magnetic field B pointing along the plane and a blue arrow for the velocity v. An orange arrow shows the force F at right angles to the plane; B points to the right and v is drawn running back into the plane. On the right a small graph shows F against the angle between v and B, a sine curve from 0 to 180 degrees with a dot marking the present angle. Step 1: v is at 90 degrees to B, F points straight down with its largest length and the dot sits at the top of the curve. Step 2: v is at 30 degrees to B, F is half as long and the dot sits halfway up the curve. Step 3: v points along B, there is no force, labelled F equals 0, and the dot is at zero. Step 4: the charge is negative, v is again at 90 degrees, and the force has the same size but points straight up.
To find the direction of , point the fingers of your right hand along and curl them towards through the smaller angle between them. Your thumb then points along . For a positive charge this is the direction of the magnetic force; for a negative charge the force is the opposite way. In Cartesian axes the cyclic products , and do the same job.
The formula also tells us how to measure . Rearranged, , so the SI unit of magnetic field is a newton per coulomb per (metre per second). This unit is the tesla (T), named after the engineer Nikola Tesla: a field pushes with a force of on a charge crossing it at right angles at . In base units,
Everyday fields are tiny in these units. The Earth's field, the one that turns a compass, is only a few times , so for weak fields people often use the older, non-SI gauss (). Table 4.2 shows how enormous the range of magnetic fields is.
| Source | Field B (tesla) |
|---|---|
| Surface of a neutron star | about |
| Strongest steady laboratory magnets | about |
| Hospital MRI scanner | to |
| Close to a small fridge magnet | about |
| Earth's field at the surface | about |
| Magnetic signals from the human brain | about |
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Answer 1.6e-14 N
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Answer (A) When it moves along the field lines
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Answer (B) Up
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Answer tesla
4.2.3 Magnetic force on a current-carrying conductor
A wire carrying a current is full of moving charges, so a magnetic field must push on the wire as a whole. Take a straight rod, length and cross-section , lying in a uniform external field (Fig. 4.4). Suppose, as in a metal, that it has one kind of mobile carrier, with charge and number density .
The rod contains carriers. In a steady current each drifts, on average, with the drift velocity of Chapter 3, so each feels a magnetic force . Adding the forces on all of them,
Here is the current density , and is the current . Let be a vector of length pointing along the rod in the direction of the current. Then , and
Notice where the direction has gone. is just a number (with a sign), so the direction of the current is carried by the vector . The size of the force is , where is the angle between the rod and the field.
3/3The forces on all carriers add up to , here pointing up with size .
A horizontal rod connected to leads at both ends lies in a region of crosses showing a magnetic field into the page; a label says into the page. Arrowheads on the leads show the current I flowing to the right. Step 1: only the rod, the field and the current are shown. Step 2: three negative charge carriers inside the rod drift to the left, shown by small blue arrows, and each has a small orange arrow pointing up for the force q v_d cross B. Step 3: the small arrows fade and one large orange arrow points up from the middle of the rod, labelled F equals I l cross B, with a double-headed arrow below the rod marking its length l.
Two cautions go with (4.4). First, is the external field, not the field that the rod's own current creates: a wire cannot push itself along. Second, the formula is for a straight piece of wire in a uniform field. A wire of any shape can be broken into short, nearly straight pieces , and the forces on them added as vectors,
a sum that becomes an integral when the pieces are made very short.
A rod held up by a magnetic field
A copper rod of mass and length hangs horizontally from two light, flexible leads. It sits in a uniform horizontal magnetic field of directed into the page, at right angles to the rod. What current must flow in the rod, and in which direction, for the leads to go slack? Take .
Try it yourself first. Then reveal the solution one step at a time.
Which way is the push?
In a vacuum chamber the magnetic field points vertically upward. Take axes with east, north and up. Find the direction of the magnetic force on (a) a proton moving east, (b) an electron moving east, (c) a neutron moving east, and (d) a proton moving straight up.
Try it yourself first. Then reveal the solution one step at a time.
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Answer 0.6 N
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Answer (A)
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Answer False
4.3 Motion in a Magnetic Field
The rate at which a force does work on a moving particle is the power . For the magnetic force, , because a cross product is perpendicular to each of the vectors that make it. So a magnetic field never does work on a charged particle and never changes its kinetic energy. It steers the particle, changing the direction of its momentum, but it cannot speed it up or slow it down. To give a particle energy you need an electric field, whose force can point along the motion; that is how every particle accelerator adds energy.
Velocity at right angles to the field
Let a particle of mass and charge move with speed in a uniform field , with perpendicular to . The magnetic force has a constant size and always points sideways, at right angles to the motion. A force of constant size that stays perpendicular to the velocity is exactly what uniform circular motion needs: the magnetic force supplies the centripetal force, and the particle moves in a circle in the plane perpendicular to (Fig. 4.5).
A particle moving in a circle of radius at speed needs a centripetal force directed towards the centre (Class XI, Motion in a Plane). Here that force is the magnetic force, of size :
Solving for the radius,
The angular speed of the particle is . Writing , where is the number of revolutions per second,
The time for one revolution is .
(4.5) says that a particle with more momentum moves on a larger circle, and that a stronger field bends it into a tighter one. (4.6(a)) holds a surprise: the speed has cancelled out. A fast particle runs round a big circle and a slow one round a small circle, but both take exactly the same time. The frequency is called the cyclotron frequency (साइक्लोट्रॉन आवृत्ति); it depends only on the particle's charge-to-mass ratio and on the field.
3/5The path closes into a circle of radius , completed in the time .
Steps 1 to 3 show a square region of crosses, a magnetic field into the page. A positive charge at the bottom moves to the right, with a blue velocity arrow and an orange force arrow pointing up towards the centre of the path. Step 1: the particle starts. Step 2: it has gone a quarter of the way round anticlockwise, its path drawn in purple; now v points up and F points left, again towards the centre. Step 3: a full circle is drawn, with the radius r marked and the formulas r equals m v over q B and T equals 2 pi m over q B beside it. Steps 4 and 5 show a side view with green field lines pointing to the right. Step 4: the velocity of the charge makes an angle of 65 degrees with the field; dashed arrows from the charge show its two components, a shorter v parallel along the field and a longer v perpendicular across it. Step 5: the path is a purple helix wound round a dashed axis along the field, with solid parts in front and dashed parts behind, the pitch p marked between two turns and the radius r marked from the axis at the right-hand end, with the label p equals v parallel T.
Velocity at an angle: the helix
Suppose instead that makes an angle with . Split the velocity into a component along the field, , and a component across it, . The magnetic force depends only on (the cross product of with a vector along is zero), so never changes. The perpendicular motion is a circle of radius , just as before. Put together, the particle drifts steadily along the field while it circles round it: its path is a helix (कुंडलिनी), like the thread of a screw.
Two numbers describe the helix. Its radius, , is the radius of the circle you would see looking straight along the field. Its pitch (चूड़ी अंतराल) is how far the particle advances along the field in one turn, like the spacing between the threads of a screw:
Reading a proton's track
In a beam line, a bending magnet with a uniform field of steers protons (mass , charge ) round a circular arc of radius , at right angles to the field. Find (a) the speed of the protons, (b) their kinetic energy in keV, taking , and (c) how long a proton would take to go once round a full circle, and how many times per second it would do so.
Try it yourself first. Then reveal the solution one step at a time.
Electrons spiralling above the poles
High above the polar regions, where the Earth's field is about , an electron () moves at in a direction making with the field. Find the radius and pitch of its helical path, and the time it takes for one turn.
Try it yourself first. Then reveal the solution one step at a time.
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Answer (B) It stays the same
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Answer False
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Answer 1.71 cm
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Answer (C) A helix wound round the field direction
4.4 Magnetic Field due to a Current Element, Biot–Savart Law
So far the magnetic field has simply been given. Now we ask how a current produces it. Only two sources of magnetic field are known: electric currents, that is, moving charges, and the intrinsic magnetism carried by particles such as electrons. This section deals with the first. The rule for currents was found in 1820 by Jean-Baptiste Biot and Félix Savart, who timed the swings of a magnetised needle at different distances from current-carrying wires.
Picture a wire carrying a steady current , cut up in imagination into tiny pieces. Choose one piece of length , and let be a vector of that length pointing along the current. The product is called a current element. We want the field that this element produces at a point P, where is the displacement from the element to P and is the angle between and (Fig. 4.7).
In vacuum, the field that a current element sets up at a point grows in proportion to and to , falls off as with the distance from the element, and points at right angles to the plane of and , along .
In symbols,
Because , the magnitude of the field is
The constant is called the permeability of free space (निर्वात की चुंबकशीलता). In SI units its value is
so .For decades this value was exact by definition. Since the 2019 revision of the SI, is a measured quantity, but it still equals to better than one part in a billion. For the direction, use the right-hand rule for a cross product: fingers along , bent towards , and the thumb gives . In Fig. 4.7 this puts the field into the page at points to the right of an upward current element and out of the page at points to its left.
1/4At right angles to the element, and is largest. points into the page.
A short orange arrow in the middle points upward, labelled I dl, with a dashed vertical line through it marking the line of the element. A bar gauge on the right shows the size of dB, and the Biot–Savart formula is written at the top left. Step 1: P is directly to the right of the element at distance r, with theta equal to 90 degrees; the field at P is drawn as a large purple cross, into the page, and the gauge is full. Step 2: P has moved round to 45 degrees above the horizontal at the same distance; the cross is smaller and the gauge is about seven tenths full. Step 3: P lies on the dashed line directly above the element; there is no field, marked dB equals 0, and the gauge is empty. Step 4: P is directly to the left of the element; the field is a large purple dot, out of the page, and the gauge is full again.
It is worth setting the Biot–Savart law beside Coulomb's law, which gives the electric field of a point charge (Table 4.3).
| Feature | Electric field of a point charge | Magnetic field of a current element |
|---|---|---|
| Source | charge , a scalar | current element , a vector |
| Dependence on distance | ||
| Direction | along , away from or towards the charge | at right angles to the plane of and |
| Dependence on angle | none: the same in every direction | ; zero along the line of the element |
| Several sources | fields add as vectors | fields add as vectors |
| Constant |
The two constants of electricity and magnetism are tied together by the speed of light in vacuum:
The match is no accident. Light is an electromagnetic wave, and Chapter 8 shows that its speed is fixed by these two constants, . Since is known exactly, and carry the same information: a laboratory that pins down one of them has, in effect, measured the other as well.
Field of a short piece of wire
A short straight piece of wire, long, lies along the -axis at the origin and carries a current of in the direction. Find the field it produces (a) at the point on the -axis, (b) at on the -axis, and (c) at a point from the origin in the -plane, on the line making with the axis.
Try it yourself first. Then reveal the solution one step at a time.
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Answer (A) A point due north of it
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Answer 1e-8 T
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Answer False
4.5 Magnetic Field on the Axis of a Circular Current Loop
The Biot–Savart law gives the field of any current once the contributions of all its elements are added up. The simplest shape for which the sum can be done exactly is a circular loop, and the simplest place to do it is a point on the loop's axis.
Take a loop of radius carrying a steady current in vacuum, with its centre O at the origin, its plane in the -plane and its axis along the -axis (Fig. 4.8). We want the field at a point P on the axis, a distance from O.
3/4The element diametrically opposite gives an equal in the opposite direction, which cancels, and an equal , which adds.
A circular loop is drawn as a tall ellipse on the left, its front half solid and its back half dashed, with arrows showing the current. The top of the loop is marked with a dot, current out of the page, and the bottom with a cross, current into the page. A horizontal axis passes through the centre O to a point P on the right, at distance x; the radius R is marked from O to the top. Step 1: a line r joins the top element d l to P, and a purple arrow d B at P is drawn at right angles to r, pointing up and to the right. Step 2: d B is split into a component d B x along the axis and a component d B perpendicular pointing up, with the angle theta between d B and the axis. Step 3: the bottom element d l prime gives a field d B prime pointing down and to the right, whose perpendicular component points down and cancels the upward one, while both axial components point to the right. Step 4: the other arrows fade and a single thick arrow B points along the axis at P, with the formula B equals mu nought I R squared over 2 times x squared plus R squared to the power three halves.
Every element of the loop is the same distance from P, . Every element is also perpendicular to the line joining it to P: an element tangent to the circle is at right angles to both the axis and the radius, and so to the plane containing them. Hence , and by (4.7(a)) the field of one element has size
is at right angles to both and , so it is tilted away from the axis. It has a component along the axis and a component at right angles to the axis.
Pair each element with the one diametrically opposite. Their perpendicular components are equal and opposite and cancel, while their axial components point the same way and add. So when we sum round the loop only survives, where is the angle between and the axis. From the geometry of the figure,
and so
Everything in this expression except is the same for every element. The lengths add up to the circumference , so the field at P is
At the centre of the loop, , and the field is
A flat coil of closely wound turns acts like loops on top of one another, so both results are multiplied by . Far out along the axis, where , the field falls off as ; we come back to this in Section 4.9.2.
The field lines of a current loop are closed curves that thread through the loop (Fig. 4.9). Their direction follows a right-hand thumb rule: curl the fingers of your right hand round the loop in the direction of the current, and your thumb points along the field inside the loop. Seen from the face where the field lines come out, the current flows anticlockwise. That face behaves like the north pole of a magnet, and the opposite face, where the current looks clockwise, like a south pole.
3/3The field lines leave one face and enter the other, just as they leave the N pole of a bar magnet and enter its S pole.
Step 1: a circular loop is drawn as an upright ellipse, its front half solid and its back half dashed, with arrows showing the current and a dashed horizontal line marking its axis. Step 2: the loop fades and its cut ends are shown, a dot at the top where the current comes out of the page and a cross at the bottom where it goes in. Blue field lines circle each cut end, and the lines near the axis pass straight through the loop from left to right, spreading out on the right and curving back round on the left; a straight line runs along the axis, labelled B. Step 3: the letter S appears on the left face of the loop and N on the right face, and a small bar magnet labelled S and N sits in the corner with the words like a magnet.
A slice of pie
A loop of wire is shaped like a slice of pie: two straight sides along radii of a circle and a circular arc of radius that subtends at the centre O (Fig. 4.10). A current of flows round it, anticlockwise as seen in the figure. (a) Find the size and direction of the magnetic field at O. (b) Keeping the same radius and current, what angle would the arc need to subtend for the field at O to be , about the size of the Earth's field?
Try it yourself first. Then reveal the solution one step at a time.
A loop shaped like a slice of pie with a right angle at its point O, bottom left. One straight side runs to the right along a radius R, the curved side is a quarter circle, and the other straight side comes back down to O. Arrowheads show the current I flowing to the right along the bottom side, anticlockwise up and round the arc, and down the left side back to O. The 90 degree angle at O is marked.
Cancelling the Earth's field
A physics laboratory needs a small region with no magnetic field. At the centre of a flat circular coil of 250 turns, mean radius , the horizontal component of the Earth's field, towards the north, is to be cancelled exactly. What current is needed, and how must the coil be placed?
Try it yourself first. Then reveal the solution one step at a time.
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Answer 1.2566 mT
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Answer (C)
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Answer (B) A south face
4.6 Ampère's Circuital Law
Adding up current elements with the Biot–Savart law always works, but the integrals can be hard. Ampère's circuital law packages the same physics differently, and whenever a current arrangement is symmetric enough it hands us the field almost for free. It does for magnetism what Gauss's law does for electrostatics.
The law is about a quantity called the circulation of round a closed path C. Imagine carrying a probe once round C, keeping a running total of how much the field points along your direction of travel: for each short step you add , which is positive where the field points the way you are going, negative where it points back, and zero where it points across your path. The total for one complete trip is written (Fig. 4.11). Ampère's law says that this total depends on nothing except the current that threads the path.
The line integral of the magnetic field round any closed curve equals times the net current passing through any surface bounded by that curve.
- the net current through the surface bounded by C, counted with the sign given by the right-hand rule
3/4Curl your right-hand fingers along C: the thumb points out of the page, so counts positive and negative. does not pierce the surface and does not count.
Three wires cross the page at right angles: I1, a dot, and I2, a cross, lie inside a green closed curve C drawn anticlockwise, and I3, a dot, lies outside it. Step 1: the curve and the three currents. Step 2: at a point on the right-hand side of C a short black arrow d l points along the curve, a purple arrow B points up and to the right, and a dashed purple line shows B t, the part of B along d l. Step 3: the labels change to plus I1 and minus I2, I3 is marked outside, and the equation line integral of B dot d l equals mu nought times I1 minus I2 is shown. Step 4: the curve is replaced by a circle of radius r centred on a single wire carrying a current I out of the page, with purple arrows for B pointing anticlockwise along the circle at four points, and the equation B times 2 pi r equals mu nought I.
Which currents count as positive? Look at C from the side from which your walk round it appears anticlockwise. Currents coming towards you count as positive and currents going away from you as negative. (This is the right-hand rule in disguise: fingers along the walk, thumb along a positive current.) Currents that do not thread C do not count at all, however close to it they pass; they change along the path, but their contributions to the circulation add up to zero.
The circulation is easy to work out when the path follows the symmetry of the field. On a stretch where runs along the path with a constant size , the contribution is simply times the length of that stretch. On stretches where crosses the path at right angles, or is zero, the contribution is zero. A path chosen so that it is made only of such stretches is called an Amperian loop (ऐम्पियरी पाश). If the stretches of the first kind have total length and the loop encloses a net current , the law becomes
Field of a long straight wire
An infinitely long straight wire carrying a current has cylindrical symmetry. Nothing changes if we turn the wire about its own axis or slide it along its length. So the field can depend only on the distance from the wire, and it circles round the wire. Choose as the Amperian loop a circle of radius centred on the wire, in a plane at right angles to it. is tangential to this circle with the same size all the way round, so and :
Three quick checks make this result believable. Distance: the field falls off as , more slowly than the of a single current element, because a long wire has elements at every distance from you. Infinity: the wire never ends, yet the field is finite everywhere off the wire; its far-off parts are distant and seen almost end-on, where is small, so they add less and less. Size: a current of in a single straight wire gives at . That is about half the Earth's field, weak by laboratory standards, yet enough to pull a compass held that close about off north.
Because the field has the same size all round any circle centred on the wire and points along that circle, the field lines are rings round the wire, which is exactly the pattern Oersted's compasses traced out. Like every magnetic field line they close on themselves, with no beginning or end; electric field lines, by contrast, start and stop on charges. The right-hand grip rule gives their sense: wrap your right hand round the wire with the thumb pointing along the current, and the curl of your fingers shows which way the field lines run.
When should you use which law? The Biot–Savart law always gives the answer, though it may need a hard integral. Ampère's law is always true, but it is useful for finding only when symmetry lets us take outside the integral: long straight wires, thick wires, solenoids, toroids. For a single circular loop no path has a constant field along it, so the field at the centre had to come from the Biot–Savart law in Section 4.5. The two laws are not rivals: Ampère's law can be proved from the Biot–Savart law, and for steady currents they contain the same physics.
A solar cable and a compass
A single straight cable carries a steady current of from a rooftop solar array to an inverter. How far from the cable must a compass be for the cable's field to fall to , roughly the size of the Earth's field? What happens to the compass reading closer than this?
Try it yourself first. Then reveal the solution one step at a time.
Now thicken the wire: let the steady current flow through a solid cylinder of radius , with the same current density everywhere in its cross-section. By symmetry the field circles round the axis, with a size that depends only on the distance from the axis. Use a circular Amperian loop of radius centred on the axis.
Outside the wire () the loop encloses the whole current, , and we recover the field of a thin wire:
Inside the wire () the loop encloses only the current flowing through the area . The current density is uniform, so
Ampère's law then gives , so
The field grows in proportion to inside the wire, reaches its largest value at the surface, and falls off as outside.
Inside a coaxial cable
A coaxial cable, like the one that brings a TV or broadband signal into a house, has a solid inner wire of radius and a thin cylindrical outer sheath of radius with the same axis. A steady current of flows along the inner wire, spread uniformly over its cross-section, and returns along the sheath. Find the magnetic field at distances (a) , (b) , (c) and (d) from the axis, and sketch against (Fig. 4.12).
Try it yourself first. Then reveal the solution one step at a time.
On the left, the cross-section of a coaxial cable: a small solid inner wire of radius a carrying current out of the page, marked with a dot, inside a thick outer ring of radius b carrying current into the page, marked with crosses. On the right, a graph of B against r: B rises in a straight line from zero at the axis to a peak at r equals a, labelled B proportional to r, then falls along a curve labelled B proportional to 1 over r until r equals b, where it drops to zero and stays at zero, labelled B equals 0.
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Answer 0.00004 T
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Answer (A)
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Answer False
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Answer (B) One half
4.7 The Solenoid
Suppose an experiment needs a magnetic field that is strong, steady and the same everywhere over a region the size of your fist. A single loop cannot give it: its field is strongest near the wire and changes from point to point. The answer is to stack many loops along a common axis by winding one long wire round a tube, turn after turn, like thread on a reel. A coil that is many times longer than it is wide is called a solenoid (परिनालिका). The wire is coated with a thin film of enamel, so neighbouring turns can touch without the current short-cutting from one turn to the next.
Fig. 4.13 shows a slice along the axis of a solenoid: the top row of dots and the bottom row of crosses are the same turns, cut through. Close to each wire the field circles round it, as it would round a straight wire. Midway between two neighbouring turns their fields point in opposite directions and cancel. Inside the solenoid, the fields of all the turns point the same way, along the axis, and add up to a strong field. Outside, the contributions of the near and far sides of each turn largely cancel, and the field is weak.
3/4For a long solenoid the field inside is uniform and along the axis, and the field outside is nearly zero.
Six turns of a solenoid are cut through along its axis: a row of dots across the top, currents out of the page, and a row of crosses along the bottom, currents into the page, with a dashed outline of the solenoid between them and the current I labelled at the top left. Step 1: a small blue circle of field surrounds each wire, anticlockwise round the dots and clockwise round the crosses. Step 2: between two neighbouring top wires two short orange arrows point in opposite directions, labelled cancel, while a green arrow inside the solenoid points to the right, labelled add. Step 3: the small circles are gone and five straight blue field lines run to the right through the inside, labelled B, with a note outside: almost no field. Step 4: a dashed green rectangle a b c d is added, with side a b of length h inside the solenoid along the field and side c d outside below the bottom row, enclosing two turns, and the equation B h equals mu nought times n h times I.
In a real solenoid of finite length some field leaks out round the ends, and the outside field is small rather than zero. The longer the solenoid, the less this leak matters. For a long solenoid we therefore use a simple model: a uniform field along the axis inside, and no field at all outside.
Let the solenoid have turns per unit length, each carrying a current . Choose the rectangular Amperian loop abcd of Fig. 4.13: side ab, of length , runs inside the solenoid parallel to the axis, and side cd runs outside it.
Along ab the field is tangential with size , so this side contributes . Along cd the field is zero. Along bc and da the field is perpendicular to the path where it is inside the solenoid and zero where it is outside, so these sides contribute nothing. Hence .
The loop encloses turns, each carrying , so . Ampère's law in the form (4.13(b)) gives
Look at what is missing from (4.16). There is no radius: a thin coil and a fat one with the same turns per metre and the same current give the same field. And there is no position: away from the ends, the field near the wall of the tube is as strong as the field on the axis. Its direction along the axis follows the right-hand rule for a loop. A field that is both strong and uniform over a large volume is exactly what MRI scanners and electron-beam experiments need, which is why the solenoid is the workhorse of magnet design. Pushing a rod of soft iron into the coil multiplies the field many times over; Chapter 5 explains how the iron joins in.
Designing a solenoid
A student needs a uniform field of inside a solenoid long, using insulated copper wire that can safely carry . (a) How many turns are needed? (b) If all the turns must fit side by side in a single layer, what is the greatest diameter the wire can have?
Try it yourself first. Then reveal the solution one step at a time.
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Answer 1.2566 mT
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Answer (B) stays the same
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Answer (C) Inside, the field is perpendicular to them; outside, it is zero
4.8 Force between Two Parallel Currents, the Ampere
In 1905 two physicists in Sydney described a copper tube taken from a lightning conductor at a coal mine in New South Wales. After a strike, part of the tube had been crushed, as if squeezed in a vice, although nothing had hit it from outside. The culprit was the lightning current itself, tens of thousands of amperes flowing along every strip of the tube's wall, with each strip pulling on the others. We can already predict this: each strip of current sits in the magnetic field of the rest, and a magnetic field pushes on a current. The cleanest case to work out is two long, straight, parallel wires.
Let wires a and b be a distance apart, carrying steady currents and in the same direction (Fig. 4.14). By (4.14), wire a produces a field of the same size at every point along wire b:
The right-hand grip rule shows that this field is perpendicular to wire b. By (4.4), a length of wire b therefore feels a force of size , directed towards wire a:
Repeating the argument with the roles swapped gives a force of the same size on a length of wire a, directed towards wire b:
3/4Wire a feels an equal and opposite force towards b, as Newton's third law requires.
Two wires cross the page at right angles, a on the left and b on the right, a distance d apart, both carrying currents out of the page, marked by dots and labelled I a and I b. Step 1: dashed blue circles show the field of wire a circling anticlockwise round it, and at wire b a blue arrow labelled B a points straight up. Step 2: an orange arrow at b points left, towards a, labelled F b a. Step 3: an equal orange arrow at a points right, towards b, labelled F a b, and the formula f equals mu nought I a I b over 2 pi d is shown. Step 4: the current in b is reversed, drawn as a cross, and both force arrows now point outwards, away from each other, with the note they repel.
Wire b is pulled towards a, and wire a towards b, by forces of equal size: for steady currents the pair obeys Newton's third law. Now reverse the current in b (Fig. 4.14, last step). The field of a at b has not changed, but now points the other way, so the force on b swings round to point away from a, and the force on a reverses too. Only the relative direction of the two currents matters: currents flowing the same way attract; currents flowing opposite ways repel. This is how the lightning current crushed the tube, since all its strips carried current the same way. Take care not to borrow the rule for charges, where the signs work the other way round.
The force per unit length, , is
The ampere
From 1946 until 2019, (4.19) was the basis of the SI unit of current. Imagine two very long, very thin, straight parallel wires placed apart in vacuum, with equal steady currents. The current in each was defined to be one ampere (ऐम्पियर) when each metre of either wire is pushed with a force of . Check this with the formula: .
Notice which unit came first. The ampere was the base unit, and the coulomb was built from it: , the charge that a steady one-ampere current carries past any cross-section in one second. Putting the definition into practice was harder than stating it, since no real wire is infinitely long. Metrologists wound precise coils, passed the same current through fixed and movable ones, and weighed the pull between them on a current balance. The forces are tiny: one ampere makes a field of only at a distance of one metre, less than one per cent of the Earth's field, so every stray field in the laboratory had to be cancelled or allowed for.
A DC power line in the Earth's field
A high-voltage DC link carries a steady current of along an overhead conductor. Take the Earth's field along the route to have a horizontal part of pointing due north, and ignore its vertical part. Find the size and direction of the magnetic force on each metre of the conductor where the line runs (a) due east, with the current flowing east, (b) due north, and (c) towards the north-east. (d) One metre of the conductor has a mass of about . Does the magnetic force matter to the engineers who string the line?
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Answer (C) Repel each other
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Answer 0.0004 N/m
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Answer (B) stays the same
4.9 Torque on Current Loop, Magnetic Dipole
4.9.1 Torque on a rectangular current loop in a uniform magnetic field
Why does the coil of an electric motor turn? A current loop in a uniform magnetic field is pushed one way on one side and the opposite way on the other. The two pushes cancel as forces, so the loop as a whole is not dragged anywhere, but they twist it. An electric dipole in a uniform electric field (Chapter 1) behaves in just the same way: no net force, but a torque. The same twist drives the meter in Section 4.10.
Take a rectangular loop carrying a current , free to turn about an axis through its centre (Fig. 4.15). Two of its sides, each of length , are parallel to the axis; the other two, each of length , are at right angles to it. The loop's area is . The field is uniform and perpendicular to the axis, and is the angle between and the normal to the loop.
2/4As the loop turns, the forces keep their size and direction, but their lines of action close up to a distance : the torque falls to .
On the left, a small perspective sketch shows a rectangular loop standing upright with a dashed vertical rotation axis through its middle, current arrows round it and green field arrows B pointing to the right above and below it. On the right, the loop is seen from above along the axis: the two sides parallel to the axis appear as a dot, current out of the page, and a cross, current into the page, joined by a thick line through the pivot, with a purple arrow for the magnetic moment m at right angles to that line. A green arrow at the bottom shows B pointing right. Step 1: the loop's plane is along the field, m points up, the dot on the left has an orange force F1 pointing up and the cross on the right has F2 pointing down; the label says tau equals I A B. Step 2: the loop has turned clockwise so that m makes an angle theta of 30 degrees with B; the forces still point up and down, and dashed lines through them are a sine theta apart; the label says tau equals I A B sine theta. Step 3: m points along B, the two forces lie along one line pulling outwards, tau equals 0, labelled stable. Step 4: m points opposite to B, the forces lie along one line pushing inwards, tau equals 0, labelled unstable.
The sides of length . These two sides carry equal currents in opposite directions through the same uniform field, so the forces on them are equal and opposite. Both forces act along the rotation axis, through the centres of these sides, so they lie on one line: they give no net force and no torque.
The sides of length . These are always at right angles to , so each feels a force of size
at right angles to , one in each direction. The two forces cancel as vectors, but they do not act along the same line, so they form a couple.
Plane of the loop along the field (). Each force acts at a distance from the axis, and the two torques add:
General angle. When the normal to the loop makes an angle with , the forces keep the same size and direction, but each line of action is now only from the axis:
Both results can be written as one vector equation. Define the magnetic moment (चुंबकीय आघूर्ण) of the loop as
where the area vector has a magnitude equal to the area of the loop and points along the normal picked out by the right-hand thumb rule. Curl your fingers along the current, and your thumb gives . The angle between and is then , and
Compare an electric dipole in an electric field, which feels : the magnetic moment plays the part of . By (4.22), its SI unit is the ampere square metre, .
The torque is zero in two positions, and (steps 3 and 4 of Fig. 4.15), but they are very different. Near , a small tilt gives a torque of about that turns the loop back: like a ball at the bottom of a bowl, this equilibrium is stable. Near , a small tilt produces a torque that increases the tilt: like a ball balanced on a hilltop, the equilibrium is unstable. Every magnetic dipole, therefore, prefers to point along the field, and this is exactly why a compass needle settles along the Earth's field.
Winding turns on the same frame multiplies every force, and so the moment, by :
Steering a small satellite
A small cube-shaped satellite turns itself with a magnetorquer: a flat square coil of 200 turns, on a side, fixed inside one face. In its orbit the Earth's field is about , and the satellite's moment of inertia about the axis it turns on is . The coil carries . (a) Find the coil's magnetic moment. (b) Find the torque on the satellite when the coil's axis is at to the field. (c) Starting from rest at , with no other torque acting, how fast is the satellite turning when the coil's axis lines up with the field? (d) Show that for small angles the satellite rocks to and fro about the field direction, and find the period of this rocking.
Try it yourself first. Then reveal the solution one step at a time.
Tipping, not spinning
A flat rectangular coil of 40 turns, , carries . It rests on a level, frictionless table with the current anticlockwise as seen from above. Take axes east, north and up. A uniform field is switched on. (a) Find the coil's magnetic moment and the torque on it. Could any uniform field, pointing in any direction, spin the coil round on the table like a turntable? (b) If the coil were free to turn in any direction, where would it come to rest, and how would its own field at its centre compare with there?
(c) The coil is replaced by a single turn of soft, flexible wire long (the perimeter of the rectangle), lying loosely on the table and carrying in the same sense, so that its moment points up. What shape does the wire take, and by what factor does its magnetic moment change compared with the same wire bent into the rectangle?
Try it yourself first. Then reveal the solution one step at a time.
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Answer 0.012 N m
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Answer True
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Answer (C) With its magnetic moment along the field
4.9.2 Circular current loop as a magnetic dipole
Step back far enough from a small current loop and its field becomes indistinguishable from that of an electric dipole. Take the axial field, (4.11). Far out, , so , and with for the area of the loop:
With this becomes, on the axis,
Now recall the field on the axis of a short electric dipole, (Chapter 1). The two formulas are the same once we translate with a small dictionary: , and . The dictionary works in other directions too. Translating the electric dipole's field on its perpendicular bisector, , gives the loop's field at a distant point in its own plane:
Shrink the loop while raising its current so that stays the same, and the approximations become exact: the limit is a point magnetic dipole. Nor does the loop have to be circular. Any flat current loop, seen from far away, produces the dipole field of its moment , and that moment is the magnetic partner of .
2/2Close up: midway between the charges the electric field (orange arrowhead) points from to , against ; inside the loop the magnetic field (orange arrowhead) points along .
Two square panels side by side, labelled electric dipole and current loop. Step 1, far away: both panels show the same pattern of blue field lines, loops that leave the centre along the axis to the right and curve back to enter it from the left, with arrows along them and a small dot at the centre for the source. Step 2, close up: the electric dipole panel shows a positive charge on the right and a negative charge on the left, with field lines starting on the plus charge and ending on the minus charge, and an orange arrow between the charges pointing left. The current loop panel shows a dot above and a cross below, the cut ends of the loop, with field lines circling each end and passing straight through the loop, and an orange arrow between them pointing right.
Close to the source the resemblance breaks down (Fig. 4.16). An electric dipole is a pair of separate charges, and its field lines begin on one and end on the other; between the charges the field points against . A current loop has no separate parts: its field lines run straight through the loop, along , and close on themselves. The magnetic version of a lone charge, a magnetic monopole, has never been found. So while an electric dipole is built from two simpler pieces, the magnetic dipole is itself the simplest magnetic object known.
Ampère pushed the analogy all the way. A current loop makes the same field as a small magnet (Fig. 4.9) and turns in a field just as a compass needle does, so perhaps every magnet is really a collection of tiny circulating currents. The idea is half right. Electrons moving round atoms do behave as tiny current loops. But an electron also carries a magnetic moment of its own, as much a part of it as its charge, and no classical picture of charge whirling round can account for it. Chapter 5 takes up this story.
How good is the dipole approximation?
A small flat coil has 100 turns of radius and carries . Find its magnetic moment, and compare the exact field on its axis from its centre with the dipole approximation. What is the field at the same distance in the plane of the coil?
Try it yourself first. Then reveal the solution one step at a time.
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Answer (C) As
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Answer False
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Answer (A) Half as big and opposite to
4.10 The Moving Coil Galvanometer
Clip a needle-type (analogue) multimeter into a torch circuit and it may read ; switch it to its voltage range and it may show across the bulb. Inside such a meter, and in every classic laboratory ammeter and voltmeter, is the same device: the moving coil galvanometer (चल कुंडली धारामापी). It turns a current into a twist, using exactly the torque on a coil that we have just worked out.
Its heart is a light rectangular coil of many turns of fine insulated wire, pivoted so that it can turn about a fixed axis (Fig. 4.17). The coil sits in the narrow gap between the curved pole pieces of a permanent magnet and a fixed cylinder of soft iron inside the coil. That arrangement makes the field in the gap strong and radial: wherever the coil's sides are, the field lines there point along a radius of the cylinder. A fine spiral hairspring resists the turning of the coil, and a light pointer fixed to the coil moves over a scale.
4/4The coil turns until the spring's restoring torque equals ; the pointer shows the deflection .
A top view of the instrument. A block labelled N on the left and a block labelled S on the right have curved faces around a grey circular soft-iron core in the middle; the coil is a thin bar through the core whose two sides sit in the curved gaps, and a small spiral spring and a pivot are at the centre. A pointer runs from the pivot down to the right, over a curved scale at the bottom. Step 1: the parts are labelled coil, core, spring, scale and pointer. Step 2: short blue arrows in both gaps point along radii, from the N pole into the core and from the core out to the S pole, showing the radial field. Step 3: a current flows, shown by a dot on the left coil side and a cross on the right; orange arrows show a force pointing up on the left side and down on the right side, turning the coil clockwise, with the label tau equals N I A B. Step 4: the coil and pointer have turned clockwise by an angle phi, marked by an arc from the dashed rest position of the pointer, and the label says k phi equals N I A B.
Because the field is radial, the plane of the coil always contains the field lines, whatever angle the coil has turned through. So in (4.21), and a coil of turns, each of area , feels a magnetic torque
The spring twists and pushes back with a torque proportional to the twist, , where is the angle of deflection. The constant measures the stiffness of the spring: it is the torque needed to twist it through one radian, and it is called the torsional constant. The coil stops at the angle where the two torques balance:
Everything in the bracket is fixed when the instrument is built, so and the scale can be marked in equal divisions. A linear scale is the main reason for making the field radial.
Sometimes we want no number at all, only a yes-or-no answer: is any current flowing, and which way? At the balance point of a Wheatstone bridge (Chapter 3) the answer should be none. A galvanometer used this way, as a current detector, is built with its zero in the middle of the scale, so a swing to the left or right shows at once which side of balance the bridge is on.
Galvanometer as an ammeter
Suppose you want to measure a current of about with a galvanometer that reaches full scale at and has a coil resistance . Two problems appear at once. The current is 500 times too big: the pointer would slam against its stop and the fine coil could burn out. And an ammeter goes in series with the circuit, so its would add to the circuit's resistance and reduce the very current you are trying to measure.
One small resistor cures both problems. A shunt (शंट), a low resistance connected in parallel with the coil, takes almost all of the current, and only a small, fixed fraction goes through the coil (Fig. 4.18). The meter's total resistance is then
Put in the numbers for our , galvanometer on a range. The shunt works out at about (the method is in the example below), so the finished ammeter adds only to the circuit: in a circuit of or more it lowers the current by 1% or less. The coil always takes the same share of the current, here, so the pointer's position tells us the total current and the scale can be marked in amperes.
2/3A small shunt in parallel carries most of the current ; only a small, fixed fraction flows through the coil.
A straight circuit line runs between two terminals, with a circle marked G and a small zigzag resistor R G next to it. Step 1: the galvanometer alone, with a thin orange arrow for the small current I G that gives full-scale deflection. Step 2: a green zigzag resistor r s is connected in parallel below G, and a dashed box round both is labelled ammeter; a thick arrow I enters, a thin arrow I G goes through G and a thick arrow I minus I G goes through the shunt. Step 3: the shunt is gone and a long purple zigzag resistor R is connected in series after G, with a dashed box round both labelled voltmeter; only the small current I G flows.
How far the pointer turns for each unit of current through the coil: .
ExampleA galvanometer that deflects for has a current sensitivity of .
Read (4.27) as a design guide: a stronger magnet, a bigger coil, a softer hairspring or more turns each raise the deflection per ampere. Each has a price. A softer spring lets the pointer wobble longer before it settles and makes the meter easy to overload, and more turns add resistance, which matters when the galvanometer becomes a voltmeter, as we see next.
A circuit with a 4.5 volt cell, a 5 ohm load resistor, and a 40 ohm galvanometer coil connected in parallel with a 0.2 ohm shunt, followed by an ideal ammeter that reads the total current. Each resistor shows the voltage across it and the current through it. You can tap a resistor to select it and change its resistance with a slider, change the cell voltage, open and close the switch, and show the moving charges.
- Read the current through the G coil and through the shunt. What fraction of the total goes through the coil? Compare it with .
- Select the shunt and raise it to its largest value, which is like removing it. What does the ammeter read now? Without any meter the current would be .
- Press the reset button to put the shunt back to , then select the load and raise it to about . Without the meter the current would now be . Is the meter's effect on the current larger or smaller than before?
Galvanometer as a voltmeter
Now suppose you want the voltage across a resistor in a circuit. A voltmeter is clipped across the resistor, in parallel, and whatever current it takes is stolen from the resistor, lowering the very voltage you want to read. Clip on a meter of resistance and the pair behaves like a single resistance ; with that is , about 1% less than . So a good voltmeter needs a resistance far larger than anything it is connected across. We get it by adding a large resistance in series with the coil, making the meter's total resistance .
How far the pointer turns for each volt across the meter: , where is the meter’s total resistance.
Is a galvanometer that is more sensitive to current also more sensitive to voltage? Not necessarily. Table 4.4 compares a coil with a rewound version that has twice as many turns of the same wire on the same frame.
| Coil | Turns | Resistance | Current sensitivity | Voltage sensitivity |
|---|---|---|---|---|
| Original | ||||
| Rewound | about (twice the wire) | (doubled) | (unchanged) |
Doubling doubles the torque per ampere, but it also doubles the length of wire, and so the resistance, and therefore the current that a given voltage can push through the coil. So what improves a galvanometer as an ammeter does not automatically improve it as a voltmeter.
Ammeter and voltmeter from one galvanometer
Ammeter
- Connected in series with the circuit
- Small shunt in parallel with the coil
- Total resistance very small, about
- Ideal ammeter: zero resistance
Voltmeter
- Connected in parallel with the component
- Large resistance in series with the coil
- Total resistance very large, about
- Ideal voltmeter: infinite resistance
Which range gives the truer reading?
A torch cell of emf and internal resistance lights a bulb of resistance . A student measures the current with a multimeter, first on its range, where the meter's resistance (galvanometer plus shunt) is , and then on its range, where it is . (a) What is the current with no meter in the circuit? (b) What does the meter read on each range, and by what percentage is each reading low? (c) What is the largest meter resistance that keeps the error below 1%?
Try it yourself first. Then reveal the solution one step at a time.
One galvanometer, two meters
A galvanometer has a coil resistance of and gives a full-scale deflection for a current of . What should be connected to it, and how, to turn it into (a) an ammeter reading up to and (b) a voltmeter reading up to ?
Try it yourself first. Then reveal the solution one step at a time.
How sensitive is a galvanometer?
A galvanometer coil has 200 turns, each of area , in a radial field of . The torsional constant of its hairspring is . Find its current sensitivity, the current that turns it through , and its full-scale current if the scale covers .
Try it yourself first. Then reveal the solution one step at a time.
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Answer (A) To make the field radial, so the torque does not depend on the coil's angle
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Answer (C) A small resistance in parallel
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Answer 4950 Ω
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Answer False
| Quantity | Symbol | SI unit | In base units | A typical value |
|---|---|---|---|---|
| Magnetic field (vector) | (tesla) | Earth's surface: about | ||
| Permeability of free space | ||||
| Magnetic moment (vector) | , the same as | satellite coil of Example 4.10: | ||
| Cyclotron frequency | proton in : | |||
| Torsional constant | galvanometer hairspring: about |
Quick recall
Moving Charges and Magnetism: key points and summary
- A current produces a magnetic field around it, as Oersted's compass showed in 1820 (Section 4.1).
- The Lorentz force on a charge is . The magnetic part is perpendicular to and does no work; its size is , and is measured in tesla (Section 4.2.2).
- A straight conductor carrying current in a uniform field feels (Section 4.2.3).
- In a uniform field a charge moves in a circle of radius at the cyclotron frequency , which does not depend on speed; with a velocity component along the path is a helix of pitch (Section 4.3).
- The Biot–Savart law gives the field of a current element: , with (Section 4.4).
- On the axis of a loop of radius the field is , which is at the centre (Section 4.5).
- Ampère's circuital law, , reduces to for a well-chosen loop, and gives for a long straight wire (Section 4.6).
- Inside a long solenoid with turns per unit length the field is uniform, , and outside it is nearly zero (Section 4.7).
- Parallel currents attract and antiparallel currents repel, with force per unit length (Section 4.8).
- A current loop has magnetic moment . In a uniform field it feels no net force but a torque , and it is in stable equilibrium with along (Section 4.9).
- Far away, a current loop is a magnetic dipole whose field matches that of an electric dipole with and (Section 4.9.2).
- In a moving coil galvanometer the torque is balanced by a spring, ; a small shunt in parallel makes it an ammeter and a large series resistance a voltmeter (Section 4.10).
Practice questions with answers (NCERT “Exercises”)
Questions 1–13 practise the same skills as the NCERT exercises, with new situations and numbers; questions 14–20 are extra practice. Take , so , and .
A science teacher wants a flat circular coil of radius that produces a field of at its centre, about three times the Earth's horizontal field, using a current of no more than . What is the smallest number of turns that will do?
Show solution
Answer64 turns (63.7 rounded up).
The number of turns must be a whole number, and 63 turns would fall just short of the required field, so at least 64 turns are needed.
A thick cable in an aluminium plant carries a steady current of . How far from the cable does its magnetic field fall to , about the strength of the Earth's field?
mShow solution
Answer.
Workers' watches and compasses within a few metres of such cables are strongly affected.
A long straight cable runs vertically up the wall of a building and carries a steady current of upward. Find the size and direction of its magnetic field at a point due east of the cable.
The size of the field, in tesla, is
TShow answer
Answer 0.000004 T
.At the point east of the cable, the field points
Show answer
Answer (A) north
With the thumb up, the fingers curl anticlockwise seen from above; on the east side that direction is north. Formally, , which is north.
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Seen from above, the field circles anticlockwise round an upward current. At a point east of the cable the anticlockwise direction is towards the north.
An underground cable lies below a pavement and carries a steady current of from south to north. Find the magnetic field it produces at the pavement directly above it. Would a compass carried along the pavement be affected?
Show solution
Answer, pointing east; a compass would be noticeably deflected.
Axes: east, north, up. The current is along and the point is above the cable, along . The field is along : it points east.
This is comparable to the Earth's horizontal field, so a compass held above the cable would swing well away from north.
A straight horizontal wire long carries in a uniform horizontal field of . At what angle to the field must the wire be turned for the force on it to be ? What is the largest force the field could exert on it, and what is the force per unit length at that setting?
°Show solution
AnswerAbout (or ); largest force , that is .
The angle gives the same force in size.
The largest force occurs with the wire at right angles to the field.
In a school current balance, a light U-shaped wire frame is pivoted so that its short end, a straight piece long, pokes into a long solenoid at right angles to the solenoid's axis. The solenoid has 3000 turns per metre and carries ; the frame carries . What is the magnetic force on the short piece, and what mass of paper rider hung on it would balance that force? Would the balance respond at all if the short piece lay along the axis instead?
NShow solution
Answer, balanced by a rider of about ; no force if the piece lies along the axis.
The force is perpendicular to both the short piece and the axis, so with the right current direction it pulls the end of the frame straight down (or pushes it up). A rider balances it when :
If the piece lay along the axis, its current would be parallel to and the force would be zero: the balance would not move.
Two parallel copper busbars in a switchboard are apart and carry each, in opposite directions. Treating the bars as long, thin, parallel wires, find the force per metre between them and the force on a length of either bar. Do they attract or repel?
NShow solution
Answer; on ; they repel.
The currents are antiparallel, so the bars repel. During a short circuit the current can be many times larger and the force grows as , so busbars must be clamped firmly.
A student winds a solenoid on a plastic pipe long and in diameter: 250 turns in a first layer, then back over it for a second layer of 250 turns, then a third layer of 250, always turning the wire the same way round the pipe. (a) Estimate the field near the centre when the coil carries . (b) A second student makes the same coil but winds the middle layer the opposite way round the pipe. What field does that coil give?
Field in the first coil, in millitesla:
mTShow answer
Answer 5.65 mT
, .Field in the second coil, in millitesla:
mTShow answer
Answer 1.885 mT
The reversed layer cancels one of the others: , so .
Show solution
The pipe's diameter does not enter: the solenoid is about 20 times longer than it is wide, so the long-solenoid formula is a good estimate near the centre.
(b) The current in the reversed layer circulates the opposite way, so its turns cancel those of one other layer. The net number of turns per metre is :
The armature of a small motor is a rectangular coil of 40 turns, , carrying between magnet poles where the field is uniform, . At one moment the plane of the coil makes with the field. Find the torque on the coil at that moment.
N mShow solution
Answer.
The plane is at to the field, so the normal (the direction of ) is at to it.
Using by mistake would give , almost three times too much. With its normal only from the field, the coil is close to a position of zero torque, so the torque is small.
A manufacturer tries three separate ways of improving a moving coil galvanometer: (i) a magnet that gives twice the field in the gap; (ii) 50% more turns of the same wire on a coil of the same area; (iii) a hairspring that is half as stiff. For each change, by what factor do the current sensitivity and the voltage sensitivity change?
(ii) More turns: the current sensitivity is multiplied by
Show answer
Answer 1.5
, so grows by 1.5.(ii) More turns: the voltage sensitivity is multiplied by
Show answer
Answer 1
The resistance also grows by 1.5, so is unchanged.Which change or changes double both sensitivities?
Show answer
Answer (C) both (i) and (iii)
Doubling or halving doubles without changing the coil's resistance, so both sensitivities double.
Show solution
(i) : doubles; is unchanged, so doubles too.
(ii) with the same wire: grows by 1.5, but the wire is 1.5 times longer, so and is unchanged.
(iii) : both and double.
Only changes that leave the coil's resistance alone improve the voltage sensitivity as well as the current sensitivity.
In a demonstration tube, electrons (, ) are accelerated from rest through a potential difference of and then enter a uniform field of at right angles to it. Why does the beam follow a circle? Find the radius of the circle.
cmShow solution
AnswerAbout .
The magnetic force is always perpendicular to the velocity and, with , has the constant size . A force of constant size that is always perpendicular to the motion keeps the speed constant and bends the path into a circle.
For the electrons of Question 11, find the frequency of revolution. The accelerating voltage is now raised to . What happens to the speed, the radius and the frequency?
The frequency of revolution, in megahertz, is
MHzShow answer
Answer 42 MHz
.With instead of , the frequency
Show answer
Answer (B) stays the same
The speed doubles (it goes as ) and so does the radius, but does not depend on speed.
Show solution
Four times the voltage gives four times the kinetic energy and so twice the speed, . The radius doubles to about , but the frequency is still : it does not depend on the speed. (At much higher speeds, close to the speed of light, this is no longer true.)
In a school demonstration motor, a flat circular coil of 25 turns, radius , carries in a uniform horizontal field of . A student holds the coil still with its normal at to the field. (a) What torque must the student apply? (b) The coil is replaced by a square coil with the same turns and current, whose area equals the circle's. Does the answer change? (c) What if the square coil has the same perimeter as the circle instead?
Torque needed, in newton metres:
N mShow answer
Answer 0.291 N m
.Square coil of the same area:
Show answer
Answer (B) same torque
depends only on the area, so the torque is the same.Square coil of the same perimeter, torque in newton metres:
N mShow answer
Answer 0.228 N m
A square of the same perimeter has of the circle's area, so the torque is .
Show solution
(b) No change: the torque depends only on , and a flat coil of any shape with the same area has the same moment.
(c) The circle's perimeter is , so the square's side is and its area , compared with for the circle. The torque falls in the same ratio, :
An electron moves due north through a region where the magnetic field points vertically upward. Which way is the magnetic force on it?
Show answer
Answer (B) West
(east); for a negative charge the force is reversed, so it points west.A uniform magnetic field can change the speed of a charged particle moving through it.
Show answer
Answer False
The force always acts at right angles to the direction of motion, so it can turn the particle but never speed it up or slow it down.What is the cyclotron frequency of a proton () in a field of , in megahertz?
MHzShow answer
Answer 18.3 MHz
.One tesla is equal to gauss.
Show answer
Answer 10000
, so .Put the steps for finding a field with Ampère's circuital law in order.
- Choose an Amperian loop along which is constant and tangential, or perpendicular, or zero
- Work out the current enclosed by the loop, with signs
- Set and solve for
- Use the symmetry to find the shape and direction of the field lines
Show answer
Answer- Use the symmetry to find the shape and direction of the field lines
- Choose an Amperian loop along which is constant and tangential, or perpendicular, or zero
- Work out the current enclosed by the loop, with signs
- Set and solve for
The loop can only be chosen well after symmetry has told you how the field lines run; then count the enclosed current and apply .A proton () enters a uniform field of with a speed of at to the field. Find the radius and the pitch of its helical path.
Show solution
AnswerRadius about ; pitch about .
The pitch is about eleven times the radius, so the helix is a long, stretched spiral.
Find your NCERT exercise question
Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.
| In your NCERT textbook | Practice question here |
|---|---|
| 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 on Moving Charges and Magnetism
- 4.1 Introduction3 questions
- 4.2.2 Magnetic field, Lorentz force4 questions
- 4.2.3 Magnetic force on a current-carrying conductor3 questions
- Velocity at an angle: the 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
- Field of a long straight wire4 questions
- 4.7 The Solenoid3 questions
- The ampere3 questions
- 4.9.1 Torque on a rectangular current loop in a uniform magnetic field3 questions
- 4.9.2 Circular current loop as a magnetic dipole3 questions
- Galvanometer as a voltmeter4 questions
Key terms (English–Hindi)
- ampereऐम्पियरaimpiyar
- The SI unit of current, formerly defined by the force of per metre between two parallel wires apart.First seen in the chapter →
- Ampère's circuital lawऐम्पियर का परिपथीय नियमaimpiyar ka paripathiya niyam
- The line integral of round any closed curve equals times the net current through it: .
- Amperian loopऐम्पियरी पाशaimpiyari paash
- A closed curve chosen so that along each part is tangential and constant, perpendicular, or zero, making Ampère's law easy to apply.First seen in the chapter →
- Biot–Savart lawबायो-सावर्ट नियमbayo-savart niyam
- The field of a current element: .
- current sensitivityधारा सुग्राहिताdhaara sugraahita
- The deflection of a galvanometer per unit current, .
- cyclotron frequencyसाइक्लोट्रॉन आवृत्तिcyclotron aavritti
- The frequency with which a charged particle circles in a uniform magnetic field; it does not depend on the particle's speed.First seen in the chapter →
- Lorentz forceलोरेंज बलlorenj bal
- The combined electric and magnetic force acting on a point charge that moves through both fields.
- magnetic momentचुंबकीय आघूर्णchumbakiya aaghurn
- For a flat coil, in ; in a field it feels a torque .First seen in the chapter →
- moving coil galvanometerचल कुंडली धारामापीchal kundali dhaaraamaapi
- An instrument in which a coil in a radial magnetic field turns against a spring through an angle proportional to the current.First seen in the chapter →
- permeability of free spaceनिर्वात की चुंबकशीलताnirvaat ki chumbakasheelta
- The constant in the laws of magnetism, playing the role that plays in electrostatics.First seen in the chapter →
- pitch (of a helix)चूड़ी अंतरालchoodi antaral
- The distance a charged particle moves along the magnetic field during one turn of its helical path, .First seen in the chapter →
- shuntशंटshunt
- A low resistance placed across a galvanometer coil, in parallel, which carries most of the current and turns the galvanometer into an ammeter.First seen in the chapter →
- solenoidपरिनालिकाparinaalika
- A long wire wound in a closely spaced helix; inside a long solenoid the field is uniform, .First seen in the chapter →
- teslaटेस्लाtesla
- The SI unit of magnetic field: . One gauss is .First seen in the chapter →
- voltage sensitivityवोल्टता सुग्राहिताvoltata sugraahita
- The deflection of a galvanometer per unit voltage across it, .