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Physics · Class 12 · Chapter 4

Moving Charges and Magnetism

Currents produce magnetic fields, and magnetic fields push on moving charges and currents. Most errors in this chapter are errors of direction, so get the cross product and the right-hand rules into your fingers before attempting numericals.

In this chapter: Lorentz force, force on a current-carrying conductor, motion of a charge in a magnetic field, Biot-Savart law, field on the axis of a circular loop, Ampere's circuital law, field of a long wire and a solenoid, force between parallel currents, torque on a current loop, the loop as a magnetic dipole, the moving coil galvanometer and its conversion to an ammeter or voltmeter.

Magnetic force on a moving charge

Oersted noticed in 1820 that a compass needle is deflected by a current in a nearby wire. Moving charges (currents) create magnetic fields, and a magnetic field exerts a force only on moving charges. For a charge q moving with velocity v in fields E and B, the Lorentz force is

F = q[E + v × B]magnetic part: F = qvB sin θ, perpendicular to both v and B
  • No magnetic force on a charge at rest, or on one moving parallel or antiparallel to B.
  • For a negative charge the force is opposite to that for a positive charge.
  • The SI unit of B is the tesla: 1 T = 1 N s C−1 m−1. A smaller unit is the gauss, 1 G = 10−4 T. The earth's field is about 3.6 × 10−5 T.

Force on a current-carrying conductor

Adding the forces on all drifting electrons in a straight wire of length l carrying current I gives

F = I l × Bl points along the current; for a wire of any shape, add I dl × B over its length

Motion of a charge in a uniform magnetic field

The magnetic force is always perpendicular to the velocity, so it does no work and cannot change the speed or kinetic energy. It only bends the path.

r = mvqBω = qBm,   T = 2πmqB,   ν = qB2πm

The time period and frequency do not depend on the speed or the radius: a faster particle moves on a bigger circle but takes the same time for one revolution. If the velocity has a component along B, that component is unaffected and the path becomes a helix. The radius uses the perpendicular component (r = mv⊥/qB) and the pitch is p = v∥T = 2πmv∥/qB.

Worked example: An electron moving at 3 × 107 m s−1 enters a field of 6 × 10−4 T at right angles. Find the radius of its path and its frequency of revolution.
Solution: r = mv/(eB) = (9 × 10−31 × 3 × 107)/(1.6 × 10−19 × 6 × 10−4) = (2.7 × 10−23)/(9.6 × 10−23) ≈ 0.28 m. Frequency ν = eB/(2πm) = (9.6 × 10−23)/(2π × 9 × 10−31) ≈ 1.7 × 107 Hz, which does not depend on the speed.

Biot-Savart law

The field produced at a point by a small current element I dl is

dB = μ04π · I dl × rr3,   |dB| = μ04π · I dl sin θr2μ0 = 4π × 10−7 T m A−1, the permeability of free space

Like Coulomb's law it is an inverse square law, but the field is perpendicular to both dl and r, and there is no field along the line of the element itself (θ = 0). The constants are linked by 1/√ε0μ0 = c.

Field on the axis of a circular loop

B = μ0IR22(x2 + R2)3/2At the centre (x = 0): B = μ0I2R   (N turns: multiply by N)

The direction follows the right-hand thumb rule for a loop: curl the fingers along the current and the thumb gives the field along the axis. Seen from one face, anticlockwise current makes that face behave like a north pole.

Ampere's circuital law

∮ B · dl = μ0Ienc

The line integral of B around any closed loop equals μ0 times the net current passing through the surface bounded by that loop. Its role is the same as Gauss's law in electrostatics: it is always true, but it gives B easily only when there is enough symmetry.

  • Long straight wire: take a circle of radius r around the wire, so B · 2πr = μ0I and B = μ0I/2πr. The field lines are circles centred on the wire. Grip the wire in your right hand with the thumb along the current; the fingers curl in the direction of B.
  • Long solenoid with n turns per unit length: taking a rectangular loop half inside and half outside gives B = μ0nI inside. The field inside is uniform and along the axis, and the field outside is nearly zero. At the ends of a long solenoid, B drops to about half this value.
Magnetic field of a straight wire and of a solenoidwww.iitmedicoguide.com(a) Long straight wireBB = μ₀I / 2πrI out of page: B anticlockwise (b) Solenoid (section)inside: B = μ₀nI (uniform)top turns: current out; bottom: inwww.iitmedicoguide.com
Field lines round a straight wire are circles whose sense follows the right-hand rule. Inside a long solenoid the lines are straight, parallel and evenly spaced, so the field is uniform; outside they spread out and return, and the field there is weak.

Force between two parallel currents

The field of wire a at wire b is μ0Ia/2πd, and wire b feels a force IbLB in that field. Per unit length:

f = μ0IaIb2πd

Parallel currents attract and antiparallel currents repel, which is the reverse of the rule for charges. For two wires 1 m apart each carrying 1 A, the force is 2 × 10−7 N per metre. This was the older definition of the ampere; since 2019 the ampere is defined by fixing the value of e, but the number is still correct.

Torque on a current loop and the magnetic dipole

A rectangular loop in a uniform field feels zero net force, but a torque acts on it. For a coil of N turns and area A carrying current I, define the magnetic moment m = NIA, with A along the normal given by the right-hand rule (unit A m2). Then

τ = m × B,   τ = NIAB sin θθ is the angle between the normal to the coil and B

The torque is maximum when the plane of the coil is parallel to B and zero when the plane is perpendicular to B. The expression is exactly like p × E for an electric dipole. Far from a small loop, the field on its axis is B = μ02m/(4πx3), the same form as the axial field of an electric dipole, so a current loop behaves as a magnetic dipole.

An electron orbiting a nucleus is a tiny current loop. Its magnetic moment is related to its orbital angular momentum by μl = (e/2me) l; the smallest value, (μl)min = eh/(4πme) = 9.27 × 10−24 A m2, is called the Bohr magneton.

Moving coil galvanometer

A coil pivoted in a radial magnetic field (made using concave pole pieces and a soft iron core) always has its plane parallel to B, so the torque is NIAB. A spring provides a restoring torque kφ, and at equilibrium

φ = NABk ICurrent sensitivity φ/I = NAB/k,   voltage sensitivity φ/V = NAB/(kRG)

Doubling the number of turns doubles the current sensitivity, but it also roughly doubles the coil resistance, so the voltage sensitivity may not change.

  • Ammeter: connect a small resistance, the shunt rs, in parallel with the galvanometer: rs = IgG/(I − Ig). An ideal ammeter has zero resistance.
  • Voltmeter: connect a large resistance R in series: R = V/Ig − G. An ideal voltmeter has infinite resistance.
Worked example: A galvanometer of resistance 12 Ω gives full-scale deflection for 3 mA. How would you convert it into an ammeter reading up to 3 A, and into a voltmeter reading up to 18 V?
Solution: Ammeter: shunt rs = (3 × 10−3 × 12)/(3 − 0.003) ≈ 0.012 Ω in parallel. Voltmeter: series resistance R = 18/(3 × 10−3) − 12 = 6000 − 12 = 5988 Ω.
Common mistakes: (1) Forgetting to reverse the force direction for an electron. (2) Saying a magnetic field can change a particle's speed; it changes only the direction. (3) Taking θ in τ = NIAB sin θ as the angle between the plane of the coil and B; it is the angle between the normal and B. (4) Using the field at the centre of a loop, μ0I/2R, for a point on the axis. (5) Connecting the shunt in series or the voltmeter resistance in parallel.

JEE and NEET focus

  • Radius, period and pitch of a charged particle in a magnetic field; comparing proton, deuteron and alpha particle.
  • Fields at the centre of loops and arcs, near straight wires and inside a solenoid, with correct directions.
  • Force per unit length between parallel wires.
  • Torque and magnetic moment of a coil; potential energy of a magnetic dipole, U = −m · B.
  • Galvanometer sensitivity and conversion to ammeter and voltmeter.

Practice questions

A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:

  1. maximum
  2. qvB
  3. zero
  4. along B
Show answer
C. F = qvB sin 0° = 0.

A proton and an alpha particle enter the same uniform field with the same velocity, at right angles to it. The ratio of their radii rp : rα is:

  1. 1 : 1
  2. 1 : 2
  3. 2 : 1
  4. 1 : 4
Show answer
B. r ∝ m/q; proton m/e, alpha 4m/2e = 2m/e.

The work done by the magnetic force on a charge moving in a circle in a uniform field is:

  1. qvB × 2πr
  2. zero
  3. positive
  4. negative
Show answer
B. The force is always perpendicular to the velocity.

The field at the centre of a circular loop of radius 10 cm carrying 5 A is about:

  1. 3.14 × 10−5 T
  2. 6.28 × 10−5 T
  3. 1.0 × 10−5 T
  4. 3.14 × 10−6 T
Show answer
A. B = μ0I/2R = (4π × 10−7 × 5)/0.2 = π × 10−5 T.

Two long parallel wires carry currents in opposite directions. They:

  1. attract
  2. repel
  3. exert no force
  4. rotate
Show answer
B. Antiparallel currents repel.

A long solenoid has 1000 turns per metre and carries 2 A. The field inside is about:

  1. 2.5 × 10−3 T
  2. 1.3 × 10−3 T
  3. 2.5 × 10−5 T
  4. 8 × 10−3 T
Show answer
A. B = μ0nI = 4π × 10−7 × 1000 × 2 ≈ 2.51 × 10−3 T.

The torque on a current loop in a uniform field is maximum when:

  1. the plane of the loop is perpendicular to B
  2. the plane of the loop is parallel to B
  3. the magnetic moment is along B
  4. the magnetic moment is opposite to B
Show answer
B. Then the normal (and m) is at 90° to B.

The period of revolution of a charged particle in a uniform magnetic field:

  1. increases with speed
  2. decreases with speed
  3. is independent of speed
  4. is proportional to the radius
Show answer
C. T = 2πm/qB.
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