By the end of this chapter you'll be able to…

  • 1Apply the Lorentz force and explain why a magnetic force can never change a particle's speed
  • 2Compute the radius, frequency and pitch for charged particles moving in a magnetic field
  • 3Use the Biot-Savart law to find the field of a straight wire and of a circular loop
  • 4Apply Ampere's circuital law by choosing an Amperian loop that exploits the symmetry
  • 5Find the field inside a solenoid from the turns per unit length
  • 6Determine the magnitude and direction of the force between two parallel currents
  • 7Compute the torque on a current loop and relate current and voltage sensitivity of a galvanometer
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Why this chapter matters
This is the chapter where electricity becomes magnetism. The Lorentz force, the Biot-Savart law and Ampere's circuital law are the tools every later magnetism question depends on, and the torque on a current loop is what makes motors and galvanometers work.

Moving Charges and Magnetism

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book. The questions run contiguously from 4.1 to 4.13 with no gaps, and a full-text search of every chapter finds zero occurrences of the phrase.

The cyclotron is promised but never taught. Older editions carried a section on motion in combined electric and magnetic fields, containing the velocity selector and the cyclotron. That section is gone: searching this chapter for "velocity selector" returns zero hits, and no cyclotron is ever described.

Yet the Introduction still tells you "We shall see how particles can be accelerated to very high energies in a cyclotron," and Summary point 3 still says the cyclotron frequency "is exploited in a machine, the cyclotron, which is used to accelerate charged particles."

So the chapter advertises a machine it no longer explains. What survives is the frequency formula itself, , developed in section 4.3 as a property of circular motion. That formula is examinable; the machine built on it is not part of this syllabus.

The toroid has been removed too. Section 4.7 is titled "The Solenoid" alone, where older editions read "The Solenoid and the Toroid." The word "toroid" does not occur anywhere in the chapter.

Textbook sectionTopic
4.1Introduction — sources of magnetic fields, and the Oersted observation
4.2Magnetic force: sources and fields, the Lorentz force, force on a conductor
4.3Motion in a magnetic field — circular and helical paths
4.4Magnetic field due to a current element, the Biot-Savart law
4.5Magnetic field on the axis of a circular current loop
4.6Ampere's circuital law
4.7The solenoid
4.8Force between two parallel currents, and the ampere
4.9Torque on a current loop, and the magnetic dipole
4.10The moving coil galvanometer

A symbol trap if you work from an extracted PDF. The permeability constant extracts as a plain "m0", the same micro-to-m substitution seen in Chapters 1 to 3. In this chapter it also hits the gauss values in Exercise 4.11.


2. The Lorentz Force (Textbook 4.2)

Oersted's observation that a current deflects a compass needle is the starting point: moving charges produce magnetic fields, and magnetic fields exert forces on moving charges.

The force on a single charge. A charge moving with velocity in a magnetic field experiences:

Combined with the electric force, this gives the full Lorentz force:

Three consequences follow from the cross product, and every one of them is examined:

  • The force vanishes when is parallel or antiparallel to , since the cross product of parallel vectors is zero.
  • The force vanishes for a stationary charge. A magnetic field does nothing to a charge at rest.
  • The force is always perpendicular to the velocity, so it does no work and cannot change the speed.

That last point is the whole content of Exercise 4.11. The magnetic force changes only the direction of motion, never the kinetic energy.

The force on a current-carrying conductor (4.2.3). A wire is a stream of moving charges, so summing the force over all of them gives:

with magnitude . The direction follows the right-hand rule or Fleming's left-hand rule.


3. Circular and Helical Motion (Textbook 4.3)

When a charge enters a field perpendicular to it, the force stays perpendicular to the velocity at every instant, which is exactly the condition for uniform circular motion. Equating the magnetic force to the centripetal requirement:

The radius grows with momentum and shrinks with field strength.

The frequency is independent of speed. From :

Neither expression contains . A faster particle traces a proportionally larger circle and completes it in the same time.

This is the result Exercise 4.12 asks you to interpret, and it is the principle the cyclotron was built on — though, as noted above, the machine itself is no longer in the chapter.

Helical motion. If the velocity has a component parallel to , that component feels no force and continues unchanged. The perpendicular component still circles, so the path is a helix. The distance advanced along the field in one revolution is the pitch:


4. The Biot-Savart Law and Its Consequences (Textbook 4.4 to 4.5)

The Biot-Savart law is to magnetism what Coulomb's law is to electrostatics: it gives the field from an infinitesimal source, to be integrated over the whole configuration.

The law. A current element produces at displacement the field:

with T m A. Note the inverse-square dependence and the cross product, which makes perpendicular to both the element and the line joining it to the point.

Straight infinite wire. Integrating along the wire gives:

at perpendicular distance . The field lines are concentric circles around the wire, with direction given by the right-hand thumb rule. This single formula answers Exercises 4.2, 4.3 and 4.4, where the only real work is deciding the direction.

On the axis of a circular loop (4.5). For a loop of radius carrying current , at axial distance :

Setting gives the field at the centre, and multiplying by turns:

which is what Exercise 4.1 needs.


5. Ampere's Circuital Law and the Solenoid (Textbook 4.6 to 4.7)

Where the Biot-Savart law always works but often demands a hard integral, Ampere's law trades generality for speed in symmetric situations — exactly as Gauss's law does in electrostatics.

The law. For any closed loop:

Only the current threading the loop counts. Currents outside contribute nothing to the integral, though they do contribute to at individual points.

Choosing the loop is the whole skill. The Amperian loop must be chosen so that is either constant along it or perpendicular to it. For a straight wire, a circle of radius makes constant and parallel to , so the integral is simply , recovering the straight-wire result in one line.

The solenoid (4.7). A long, closely wound solenoid has a nearly uniform field inside and a negligible field outside. A rectangular Amperian loop straddling the winding gives:

where is the number of turns per unit length — not the total number of turns. Converting the total turns and the length into is the step Exercises 4.6 and 4.8 are really testing.

For the multi-layer solenoid of Exercise 4.8, all five layers count: is the total turns across every layer divided by the length. The diameter is given but never needed, since inside does not depend on the radius.


6. Force Between Parallel Currents (Textbook 4.8)

Each wire sits in the magnetic field of the other, so each feels a force. Combining the straight-wire field with the force on a conductor gives the force per unit length:

The direction rule is worth memorising exactly, because it runs opposite to the electrostatic case that students meet first:

  • Parallel currents, flowing the same way, attract.
  • Antiparallel currents, flowing opposite ways, repel.

Two like charges repel; two like currents attract. Exercise 4.7 turns on this distinction, and Exercise 4.5 uses the same formula in its simplest form.

This relation defines the ampere. Historically, one ampere was the current which, flowing in two infinitely long parallel wires one metre apart in vacuum, produces a force of N per metre of length. The 2019 SI revision redefined the ampere through a fixed value of the elementary charge, but this force relation remains how the unit is taught here.


7. Torque on a Current Loop, and the Galvanometer (Textbook 4.9 to 4.10)

A current loop is a magnetic dipole. In a uniform field, the forces on opposite sides of a loop are equal and opposite, so the net force is zero — but they do not act along the same line, so they produce a torque.

Defining the magnetic moment of a loop of turns, area , carrying current :

with magnitude , where is the angle between the normal to the loop and the field.

Two facts fall straight out of this, and Exercise 4.13 asks about both:

  • The torque is maximum when the plane of the coil is parallel to the field, and zero when the plane is perpendicular to it.
  • The torque depends only on the enclosed area, not on the shape. A circular, square or irregular loop of equal area experiences the same torque.

A circular loop as a magnetic dipole (4.9.2). Comparing the axial field of a loop at large distances with the field of an electric dipole shows the same falloff, which is what justifies calling the loop a dipole at all. This is the bridge into Chapter 5.

The moving coil galvanometer (4.10). A coil suspended in a radial field turns until the magnetic torque balances the restoring torque of the spring, , so the deflection is proportional to the current.

Raising current sensitivity need not raise voltage sensitivity. If more turns are added, rises but so does the resistance of the longer wire, and the two effects can cancel exactly. Exercise 4.10 is built on precisely this: the current sensitivity ratio is 1.4 while the voltage sensitivity ratio is 1.0.


Summary

  • , and with the electric term the Lorentz force is .
  • The magnetic force is always perpendicular to , so it does no work and never changes the speed — only the direction.
  • No magnetic force acts on a stationary charge, or on one moving parallel to the field.
  • On a conductor, , with magnitude .
  • Perpendicular entry gives circular motion with ; the frequency is independent of speed.
  • A parallel velocity component makes the path a helix of pitch .
  • Biot-Savart: , with T m A.
  • Straight wire: . Circular loop centre: . On the axis: .
  • Ampere's law counts only the enclosed current, and is quick only when the loop is chosen to exploit symmetry.
  • Solenoid: with the turns per unit length; the radius does not enter.
  • Parallel currents attract and antiparallel currents repel, with — the opposite convention to like charges.
  • A current loop is a magnetic dipole of moment , feeling zero net force but a torque in a uniform field.
  • Torque depends on the enclosed area alone, not the shape of the loop.
  • Galvanometer: current sensitivity , voltage sensitivity — raising one need not raise the other.
  • The cyclotron and velocity selector have been removed from this chapter, though the Introduction and Summary still refer to the cyclotron; only the frequency formula remains.
  • The toroid has been removed; section 4.7 now covers the solenoid alone.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Lorentz force on a charge
F = q(E + v x B)
The magnetic part is perpendicular to v, so it does no work and cannot change the speed
Magnetic force magnitude
F = qvB sin(theta)
Zero when v is parallel to B, and zero for a stationary charge
Force on a current-carrying conductor
F = I(l x B), with magnitude BIl sin(theta)
Obtained by summing the Lorentz force over all the moving charges in the wire
Radius of the circular path
r = mv/(qB)
The radius grows with momentum and falls with field strength
Cyclotron frequency
nu = qB/(2 pi m), and omega = qB/m
Independent of speed and radius; a faster particle traces a bigger circle in the same time
Pitch of the helical path
p = 2 pi m v_parallel/(qB)
The velocity component along B feels no force and advances the helix
Biot-Savart law
dB = (mu_0/4 pi) I dl x r-hat / r squared
mu_0 = 4 pi x 10^-7 T m A^-1; the field is perpendicular to both the element and the line to the point
Field of a long straight wire
B = mu_0 I/(2 pi a)
Field lines are concentric circles; direction by the right-hand thumb rule
Field on the axis of a circular loop
B = mu_0 I R squared / [2(R squared + x squared)^(3/2)]
Setting x = 0 gives the centre value
Field at the centre of a coil
B = mu_0 N I/(2R)
N is the total number of turns of the coil
Ampere's circuital law
The line integral of B around a closed loop equals mu_0 times the enclosed current
Only current threading the loop counts; the loop must be chosen to exploit symmetry
Field inside a solenoid
B = mu_0 n I
n is turns per unit length, not the total turns; the radius does not enter
Force between parallel currents
F/l = mu_0 I1 I2/(2 pi d)
Parallel currents attract, antiparallel currents repel, the opposite of like charges
Magnetic moment of a current loop
m = N I A, directed along the normal to the loop
This is what makes a current loop behave as a magnetic dipole
Torque on a current loop
torque = m x B, with magnitude N I A B sin(theta)
Net force is zero in a uniform field, but the torque is not; theta is measured from the normal
Galvanometer sensitivities
Current sensitivity = NAB/k; voltage sensitivity = NAB/(kR)
Adding turns raises N but also R, so raising current sensitivity need not raise voltage sensitivity
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Assuming a magnetic field can speed a charged particle up
The magnetic force is always perpendicular to the velocity, so it does zero work. The speed and kinetic energy are unchanged; only the direction of motion turns. This is the entire point of Exercise 4.11.
WATCH OUT
Using the total number of turns as n in B = mu_0 n I for a solenoid
n is the number of turns per unit length. Divide the total turns by the length in metres first. In Exercise 4.8 all five layers must be counted before dividing by 0.8 m.
WATCH OUT
Expecting parallel currents to repel because like charges repel
Currents follow the opposite convention: parallel currents attract and antiparallel currents repel. Deriving it once from F = I(l x B) fixes it better than memorising.
WATCH OUT
Measuring the torque angle from the plane of the coil instead of the normal
In torque = N I A B sin(theta), theta is the angle between the field and the NORMAL to the coil. Torque is maximum when the plane is parallel to B and zero when the plane is perpendicular to it.
WATCH OUT
Thinking the torque depends on the shape of the loop
Only the enclosed area matters. A circular, square or irregular loop of the same area in the same field with the same current feels the same torque, as Exercise 4.13(b) makes explicit.
WATCH OUT
Assuming a more sensitive galvanometer in current is also more sensitive in voltage
Voltage sensitivity is NAB/(kR). Increasing the number of turns increases N but also increases the resistance R, and the effects can cancel. In Exercise 4.10 the current sensitivity ratio is 1.4 while the voltage sensitivity ratio is exactly 1.0.
WATCH OUT
Believing Ampere's law says B is zero wherever the enclosed current is zero
The law constrains only the line INTEGRAL of B around the loop. External currents contribute nothing to that integral but do contribute to B at individual points on it.
WATCH OUT
Using the solenoid diameter in the field calculation
B = mu_0 n I inside a long solenoid is independent of the radius. Exercise 4.8 supplies the diameter precisely so that you must recognise it is not needed.
WATCH OUT
Studying the cyclotron as machinery because the Introduction mentions it
The section on motion in combined electric and magnetic fields, containing the velocity selector and the cyclotron, has been removed. Only the frequency formula nu = qB/2 pi m survives, in section 4.3.
WATCH OUT
Dropping the direction when a question asks for the magnetic field
Exercises 4.3 and 4.4 are almost entirely about direction. State the direction explicitly using the right-hand thumb rule, since the magnitude alone will not score full marks.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Moving Charges and Magnetism?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • F = q(E + v x B); the magnetic part does no work and never changes the speed
  • No magnetic force acts on a stationary charge or one moving parallel to B
  • On a conductor, F = I(l x B), magnitude BIl sin(theta)
  • Perpendicular entry gives a circle of radius r = mv/(qB)
  • The frequency nu = qB/(2 pi m) is independent of speed and radius
  • A parallel velocity component makes the path a helix of pitch 2 pi m v_parallel/(qB)
  • Biot-Savart gives dB from a current element, with mu_0 = 4 pi x 10^-7 T m A^-1
  • Straight wire: B = mu_0 I/(2 pi a), with circular field lines
  • Circular loop centre: B = mu_0 N I/(2R); on the axis the denominator carries the 3/2 power
  • Ampere's law counts only the enclosed current and is quick only when the loop matches the symmetry
  • Solenoid: B = mu_0 n I with n the turns per unit length; the radius does not enter
  • Parallel currents attract, antiparallel repel, with F/l = mu_0 I1 I2/(2 pi d)
  • A current loop is a magnetic dipole of moment m = N I A
  • Torque = m x B; net force is zero in a uniform field but the torque is not
  • Torque depends on enclosed area alone, not on the shape of the loop
  • Galvanometer current sensitivity NAB/k, voltage sensitivity NAB/(kR)
  • The cyclotron, velocity selector and toroid have been removed from this chapter
  • The Additional Exercises block has been removed, leaving Exercises 4.1 to 4.13

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit III: Magnetic Effects of Current and Magnetism, no chapter-wise split published by CBSE

Question typeMarks eachTypical countWhat it tests
Lorentz Force, Force on a Current-Carrying Conductor and Motion in a Magnetic Field2-31Force on charges and conductors, circular and helical paths, radius and frequency
Biot-Savart Law, Field of a Straight Wire, Field at the Centre of a Coil, Ampere's Circuital Law and the Solenoid3-41Fields from currents, choice of Amperian loop, and turns per unit length
Force Between Parallel Currents, Torque on a Loop and the Galvanometer4-51Attraction and repulsion of currents, magnetic moment, torque, and meter sensitivity
Prep strategy
  • Draw the vectors before computing any cross product, and state the direction in words as well as symbols
  • Convert total turns into turns per unit length the moment a solenoid appears
  • Check whether the question wants the angle from the plane or from the normal before using sin(theta)
  • Derive the parallel-current direction rule once rather than memorising it, since it inverts the electrostatic habit
  • Remember that a magnetic force does no work, so any question about speed or kinetic energy has a one-line answer

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Electric motors

The torque on a current loop in a magnetic field is exactly the principle a motor runs on, with a commutator reversing the current each half turn to keep the rotation going.

Moving coil meters

Analogue ammeters and voltmeters are galvanometers with shunts or multipliers, reading current through the deflection NIAB = k phi.

Mass spectrometers

Since r = mv/qB, ions of different mass follow different radii in the same field, separating isotopes along a detector.

Magnetic resonance imaging

Large superconducting solenoids produce the uniform fields that MRI depends on, using the same B = mu_0 n I relation scaled up.

Electromagnets and relays

A solenoid wound on a soft iron core converts a small control current into a strong switchable field, operating contacts in relays and starters.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Sketch the field direction with the right-hand thumb rule before writing any magnitude
2
Write down whether the given angle is from the plane or from the normal, in words, before substituting
3
Convert every length to metres and every gauss to tesla in the first line of working
4
For solenoids, compute n = total turns divided by length as a separate labelled step
5
State attraction or repulsion explicitly in parallel-current questions; the magnitude alone is incomplete
6
When asked about speed or kinetic energy in a magnetic field, answer from the zero-work argument rather than computing
7
Check whether a supplied datum such as a diameter is actually needed, since some are deliberately redundant

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
The Hall effect, where a transverse voltage appears across a current-carrying conductor in a magnetic field and reveals the sign of the carriers
STRETCH
Magnetic vector potential, from which B is obtained as a curl and which generalises the Biot-Savart law
STRETCH
The relativistic origin of magnetism, in which the magnetic force on a moving charge is the electric force seen from another frame
STRETCH
Helmholtz coils, a pair of separated loops arranged to give an unusually uniform field near their common axis
🚀

JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainSpeed independence of the cyclotron frequencyCircular motion in a magnetic field

An electron enters a uniform field G perpendicular to its velocity of m s. Find the radius and the frequency of revolution, and state whether either depends on the speed.

Stuck? Show the approach

Equate the magnetic force to the centripetal requirement, then form the frequency and check which variables survive.

Show the full solution

With G T, T. From , m. The frequency is Hz. The radius is proportional to , but contains no at all, because a faster electron travels a proportionally larger circle in the same period.

Answer: r = 4.2 x 10^-2 m; nu = 1.8 x 10^7 Hz; the radius depends on speed but the frequency does not
The trap

Forgetting that a gauss is 10^-4 tesla, or assuming a faster particle must circulate more often.

JEE MainCurrent versus voltage sensitivityMoving coil galvanometer

Two galvanometers have , , m, T and , , m, T, with identical spring constants. Find the ratio of the current sensitivity and of the voltage sensitivity of the second to the first.

Stuck? Show the approach

Form the ratio NAB/k for current sensitivity, then divide each by its own resistance for voltage sensitivity.

Show the full solution

Current sensitivity is , and the spring constants cancel because they are equal. The ratio is therefore . Voltage sensitivity is , which carries an extra factor of , so its ratio is the current-sensitivity ratio multiplied by . That gives , since the higher resistance of the second meter exactly offsets its greater current sensitivity.

Answer: Current sensitivity ratio 1.4; voltage sensitivity ratio 1.0
The trap

Assuming the more current-sensitive meter is automatically the more voltage-sensitive one. The added turns raise the resistance too, and here the two effects cancel exactly.

JEE AdvancedTorque is set by area, not shapeMagnetic dipole in a uniform field

A circular coil of 30 turns and radius 8.0 cm carries 6.0 A in a uniform field of 1.0 T with the field in the plane of the coil. Find the counter torque needed to prevent it turning, and state whether the answer changes if the coil is made irregular but keeps the same area.

Stuck? Show the approach

Use torque = N I A B sin(theta) with theta measured from the normal, then consider what the derivation actually used.

Show the full solution

The field lies in the plane of the coil, so the normal is at to and . The area is m, giving N m. The derivation used only the enclosed area, never the boundary shape, so an irregular loop of the same area feels the same torque.

Answer: 3.1 N m; the answer is unchanged for any shape of the same area
The trap

Assuming the shape matters. The magnetic moment is NIA, and A is the enclosed area whatever the outline.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardHigh
JEE MainHigh
NEETHigh
JEE AdvancedMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because the magnetic force q(v x B) is a cross product, and a cross product is always perpendicular to both of its factors. In particular it is perpendicular to the velocity. Work is the component of force along the displacement, and a perpendicular force has no such component, so the work done is exactly zero at every instant. With no work done, the kinetic energy and therefore the speed cannot change. The direction of motion changes continuously, which is why the path curves.

They are different phenomena and there is no reason for them to agree. Take the field of wire one at the position of wire two: it is perpendicular to wire two, and applying F = I(l x B) gives a force pointing back towards wire one when the currents run the same way. Reversing one current reverses the force. The reliable approach is to work it out from the field and the force law once, rather than trying to import intuition from electrostatics.

Biot-Savart always works but often requires a difficult integral. Ampere's law is faster, but only when the symmetry lets you choose a loop along which B is either constant in magnitude and parallel to the path, or perpendicular to it. Long straight wires and long solenoids satisfy this; a finite loop viewed off-axis does not. The relationship mirrors that between Coulomb's law and Gauss's law in electrostatics.

No. For a long, closely wound solenoid the field inside is uniform and equal to mu_0 n I, which contains no radius. Exercise 4.8 supplies the diameter deliberately, and part of what it tests is whether you recognise that the datum is not needed. What does matter is n, the turns per unit length, so remember to count every layer of winding and divide by the length in metres.

The section on motion in combined electric and magnetic fields, which contained the velocity selector and the cyclotron, has been removed from this edition. However the chapter's Introduction still says particles will be accelerated in a cyclotron, and the Summary still refers to the machine. What remains examinable is the cyclotron frequency nu = qB/(2 pi m), which is derived in section 4.3 as a property of circular motion. Learn the formula and its speed independence; do not spend time on the machine's construction.

The forces on opposite sides of the loop are equal in magnitude and opposite in direction, so they cancel when added. They do not, however, act along the same line of action, so they form a couple and produce a torque. This is what makes a current loop behave like a magnetic dipole: it tends to rotate until its magnetic moment aligns with the field, but it does not translate. In a non-uniform field the cancellation fails and a net force appears.

In torque = N I A B sin(theta) the angle is always between the magnetic moment, which points along the normal to the loop, and the field. If a question says the field lies in the plane of the coil, then the normal is perpendicular to the field, theta is 90 degrees, and the torque is maximum. If the field is perpendicular to the plane, the normal is along the field, theta is zero, and the torque vanishes. Sketching the normal before substituting removes the ambiguity.
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Last reviewed on 18 August 2026. Written and reviewed by subject-matter experts — read about our process.
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