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CBSE · Class 12 · Physics

Chapter 4: Moving Charges and Magnetism

7 concepts · 56 practice questions with answers · Free to read

Magnetic force on a moving charge and on a current

From daily life

The picture tube of an old television bends the electron beam with magnets; an electric motor turns because wires in a magnetic field feel a force. Link to earlier classes : the force on a current-carrying wire in a magnetic field from Class 10.

Simple definition

A charge moving in a magnetic field B feels the Lorentz force, which is perpendicular to both its velocity and the field. A charge at rest feels no magnetic force. The SI unit of B is the tesla (T).

Rule 1: Force on a moving charge — F = q v B sin θ (vector form F = q v × B), where θ is the angle between v and B. With an electric field as well, F = q (E + v × B). The magnetic force does no work, because it is perpendicular to the velocity.

  • Easy : q = 2 μC moving at 3 × 10⁵ m s⁻¹ at right angles to B = 0.5 T ⇒ F = 2 × 10⁻⁶ × 3 × 10⁵ × 0.5 = 0.30 N.
  • Medium : A proton at 2 × 10⁶ m s⁻¹ at 30° to B = 0.2 T feels F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.2 × 0.5 = 3.2 × 10⁻¹⁴ N.
  • Exam : In a velocity selector the forces balance when qE = qvB, so v = E ÷ B. For E = 2 × 10⁴ V m⁻¹ and B = 0.10 T, v = 2 × 10⁵ m s⁻¹.

Rule 2: Force on a current-carrying wire — F = I l B sin θ (I = current, l = length of wire in the field). The direction is given by Fleming's left-hand rule.

  • Easy : l = 0.5 m, I = 4 A, B = 0.3 T at 90° ⇒ F = 0.6 N.
  • Medium : The same wire at 30° to the field feels 0.6 × sin 30° = 0.3 N.
  • Exam : If the wire is parallel to the field (θ = 0°), the force is zero.
Remember

F = qvB sin θ; F ⟂ v and F ⟂ B; magnetic force does no work.

Common mistake

Writing that a magnetic field speeds up a charge. It can only change the direction of the velocity.

Shortcut

Memory clue : "Left hand: First finger field, seCond finger current, thuMb motion (force)."

Spot the difference

ForceActs onDepends on speed?Does work?
Electricany chargenoyes
Magneticmoving chargeyesno

Practice questions

Easy
  1. 1What is the magnetic force on a charge at rest?
  2. 2What is the direction of the Lorentz force with respect to v and B?
  3. 3How much work does the magnetic force do on a moving charge?
Medium
  1. 4Find the force on a 2 μC charge moving at 3 × 10⁵ m s⁻¹ perpendicular to a 0.5 T field.
  2. 5Find the force on a 0.5 m wire carrying 4 A at right angles to a 0.3 T field.
  3. 6Find the speed selected by E = 2 × 10⁴ V m⁻¹ and B = 0.10 T.
Challenging
  1. 7Find the force on the wire of question 5 when it makes 30° with the field.
  2. 8An electron moves at 3 × 10⁶ m s⁻¹ perpendicular to a 0.4 T field. Find the force on it.
Show answers (try them yourself first)
  1. 1Zero.
  2. 2It is perpendicular to both v and B.
  3. 3Zero.
  4. 4F = qvB = 0.30 N.
  5. 5F = IlB = 0.6 N.
  6. 6v = E ÷ B = 2 × 10⁵ m s⁻¹.
  7. 7F = 0.6 × sin 30° = 0.3 N.
  8. 8F = 1.6 × 10⁻¹⁹ × 3 × 10⁶ × 0.4 = 1.9 × 10⁻¹³ N.

Motion of a charged particle in a magnetic field

From daily life

Particle accelerators and mass spectrometers bend beams of charged particles into circles using magnets. Link to earlier classes : circular motion and centripetal force from Class 11.

Simple definition

When a charged particle enters a uniform magnetic field at right angles to it, the magnetic force acts as the centripetal force and the particle moves in a circle at constant speed. If its velocity has a component along the field as well, the path is a helix.

Rule 1: Circular motion — r = m v ÷ (q B) and T = 2π m ÷ (q B) (m = mass, q = charge, v = speed). The period and frequency do not depend on the speed or the radius.

  • Easy : A proton at 2 × 10⁶ m s⁻¹ in B = 0.20 T moves in a circle of r = 1.67 × 10⁻²⁷ × 2 × 10⁶ ÷ (1.6 × 10⁻¹⁹ × 0.20) = 0.104 m.
  • Medium : Its period is T = 2π × 1.67 × 10⁻²⁷ ÷ (1.6 × 10⁻¹⁹ × 0.20) = 3.3 × 10⁻⁷ s, so the frequency is about 3.0 MHz.
  • Exam : An electron accelerated through 100 V reaches 5.9 × 10⁶ m s⁻¹; in B = 1.0 mT it follows a circle of r = 9.1 × 10⁻³¹ × 5.9 × 10⁶ ÷ (1.6 × 10⁻¹⁹ × 10⁻³) = 3.4 cm.

Rule 2: Helical motion — Split v into v∥ (along B, unchanged) and v⊥ (across B, giving a circle). The pitch is p = v∥ × T, the distance advanced along the field in one turn.

  • Easy : An electron with v = 3 × 10⁶ m s⁻¹ at 60° to B = 0.010 T has v⊥ = 2.6 × 10⁶ and v∥ = 1.5 × 10⁶ m s⁻¹.
  • Medium : Its radius is r = m v⊥ ÷ (eB) = 1.5 mm and its period is T = 3.6 × 10⁻⁹ s.
  • Exam : The pitch is v∥ T = 1.5 × 10⁶ × 3.6 × 10⁻⁹ = 5.4 mm.
A positive charge moving in a circle in a magnetic field into the page××××××××××××××××××××××××+qvFrB into the page

In the figure : A positive charge entering a field B (into the page) moves on a circle; the force F points to the centre.

Remember

r = mv/qB; T = 2πm/qB (independent of v); the kinetic energy stays constant.

Common mistake

Saying that a faster particle takes a shorter time per turn. The radius grows with speed, so the period stays the same.

Shortcut

Memory clue : "Radius grows with momentum (mv over qB); the time for a turn is fixed."

Spot the difference

Angle between v and BPath
0° or 180°straight line
90°circle
any other anglehelix

Practice questions

Easy
  1. 1What is the path of a charge entering a magnetic field at 90° to it?
  2. 2Does the period of the circular motion depend on the speed?
  3. 3Does the kinetic energy of the particle change in the field?
Medium
  1. 4Find the radius for a proton at 2 × 10⁶ m s⁻¹ in a 0.20 T field.
  2. 5Find the period for the proton of question 4.
  3. 6Find the radius for an electron accelerated through 100 V entering a 1.0 mT field.
Challenging
  1. 7An electron with v = 3 × 10⁶ m s⁻¹ makes 60° with a 0.010 T field. Find the radius and the pitch of its helix.
  2. 8Why can a magnetic field not change the kinetic energy of a charged particle?
Show answers (try them yourself first)
  1. 1A circle.
  2. 2No.
  3. 3No.
  4. 4r = mv ÷ qB ≈ 0.104 m.
  5. 5T = 2πm ÷ qB ≈ 3.3 × 10⁻⁷ s.
  6. 6v = 5.9 × 10⁶ m s⁻¹, so r ≈ 3.4 cm.
  7. 7r ≈ 1.5 mm; pitch ≈ 5.4 mm.
  8. 8The force is perpendicular to the velocity, so it does no work.

Biot–Savart law and the field of a circular loop

From daily life

An electromagnet used in a crane and the coil in a speaker both produce a field through current in wires. Link to earlier classes : the magnetic effect of current from Class 10.

Simple definition

The Biot–Savart law gives the magnetic field produced by a small current element. It is the magnetic counterpart of Coulomb's law. The field direction is found by the right-hand thumb rule.

Rule 1: Biot–Savart law — dB = (μ₀ ÷ 4π) × I dl sin θ ÷ r², where μ₀ = 4π × 10⁻⁷ T m A⁻¹ is the permeability of free space (r = distance from the element, θ = angle between dl and r).

  • Easy : μ₀ ÷ 4π = 10⁻⁷ T m A⁻¹.
  • Medium : The field at a point along the line of the element (θ = 0°) is zero.
  • Exam : The field varies as 1 ÷ r², like the electric field of a point charge, but its direction is perpendicular to the plane of dl and r.

Rule 2: Circular loop — At the centre, B = μ₀ N I ÷ (2R). On the axis at a distance x, B = μ₀ N I R² ÷ [2 (R² + x²)^(3/2)] (R = radius, N = number of turns).

  • Easy : I = 5 A, R = 0.10 m, N = 1 ⇒ B(centre) = 4π × 10⁻⁷ × 5 ÷ 0.20 = 3.14 × 10⁻⁵ T.
  • Medium : With N = 100 turns the field is 3.14 × 10⁻³ T.
  • Exam : At x = R on the axis, B = B(centre) ÷ 2^(3/2) = 0.354 × B(centre); for the single loop above it is 1.11 × 10⁻⁵ T.
Remember

μ₀ = 4π × 10⁻⁷ T m A⁻¹; B(centre) = μ₀NI/2R; the field is strongest at the centre of the loop.

Common mistake

Using the centre formula at points on the axis away from the centre.

Shortcut

Memory clue : "Curl the fingers along the current, the thumb points along the field."

Spot the difference

LawSourceFalls as
Coulombpoint charge1/r²
Biot–Savartcurrent element1/r², direction ⟂

Practice questions

Easy
  1. 1Write the value of μ₀ ÷ 4π.
  2. 2Which rule gives the direction of the field around a wire?
  3. 3What is the field at a point on the line of a current element?
Medium
  1. 4Find the field at the centre of a single loop of radius 0.10 m carrying 5 A.
  2. 5Find the field at the centre if the coil has 100 turns.
  3. 6Find the field on the axis of the single loop at x = 0.10 m.
Challenging
  1. 7What is the ratio of the field at x = R on the axis to that at the centre?
  2. 8A coil of 50 turns, radius 0.10 m and current 5 A is given. Find the field at its centre.
Show answers (try them yourself first)
  1. 110⁻⁷ T m A⁻¹.
  2. 2The right-hand thumb rule.
  3. 3Zero, since sin θ = 0.
  4. 4B = μ₀I ÷ 2R = 3.14 × 10⁻⁵ T.
  5. 5B = 3.14 × 10⁻³ T.
  6. 6B = 1.11 × 10⁻⁵ T.
  7. 71 ÷ 2^(3/2) = 0.354.
  8. 8B = μ₀NI ÷ 2R = 1.57 × 10⁻³ T.

Ampere's circuital law and the solenoid

From daily life

A door bell and a relay use a coil of many turns to make a strong, steady magnet. Link to earlier classes : the field of a solenoid from Class 10.

Simple definition

Ampere's circuital law says that the line integral of B around any closed loop equals μ₀ times the current passing through the loop. It makes the field of symmetric currents easy to find, the way Gauss's law does for charges.

Rule 1: Straight wire — ∮ B · dl = μ₀ I, which gives B = μ₀ I ÷ (2π r) at a distance r from a long straight wire. The field lines are circles around the wire.

  • Easy : I = 10 A and r = 0.05 m ⇒ B = 2 × 10⁻⁷ × 10 ÷ 0.05 = 4 × 10⁻⁵ T.
  • Medium : I = 30 A and r = 0.20 m ⇒ B = 3 × 10⁻⁵ T.
  • Exam : If the distance is doubled the field halves (B ∝ 1 ÷ r).

Rule 2: Solenoid — Inside a long solenoid, B = μ₀ n I (n = number of turns per unit length). The field is uniform inside and almost zero outside.

  • Easy : 500 turns on 0.25 m with I = 2 A gives n = 2000 m⁻¹ and B = 4π × 10⁻⁷ × 2000 × 2 = 5.0 × 10⁻³ T.
  • Medium : n = 3000 m⁻¹ and I = 1.5 A ⇒ B = 5.7 × 10⁻³ T.
  • Exam : Doubling the current while halving n leaves B unchanged. An iron core raises B many times.
Solenoid with a uniform field inside and closed field lines outsideB inside (uniform)coil of n turns per unit length carrying I

In the figure : A solenoid: uniform field inside, field lines closing outside like those of a bar magnet.

Remember

B = μ₀I/2πr near a wire; B = μ₀nI inside a solenoid.

Common mistake

Using the total number of turns N instead of the turns per metre n.

Shortcut

Memory clue : "Solenoid: mu-nought n I; wire: mu-nought I over two pi r."

Spot the difference

SourceField
Long straight wireμ₀I ÷ 2πr, circular lines
Solenoid (inside)μ₀nI, uniform and axial

Practice questions

Easy
  1. 1State Ampere's circuital law.
  2. 2Write the formula for the field inside a long solenoid.
  3. 3What is the field outside an ideal long solenoid?
Medium
  1. 4Find the field at 5 cm from a long wire carrying 10 A.
  2. 5Find the field inside a solenoid of 500 turns on 0.25 m carrying 2 A.
  3. 6Find the field at 0.20 m from a wire carrying 30 A.
Challenging
  1. 7Find the field inside a solenoid with 3000 turns per metre carrying 1.5 A.
  2. 8What happens to B inside a solenoid when the current is doubled and n is halved?
Show answers (try them yourself first)
  1. 1The line integral of B around a closed loop is μ₀ times the current enclosed.
  2. 2B = μ₀ n I.
  3. 3Practically zero.
  4. 4B = 4 × 10⁻⁵ T.
  5. 5B = 5.0 × 10⁻³ T.
  6. 6B = 3 × 10⁻⁵ T.
  7. 7B = 4π × 10⁻⁷ × 3000 × 1.5 ≈ 5.7 × 10⁻³ T.
  8. 8It remains unchanged, because nI is the same.

Force between two parallel currents

From daily life

Two parallel power cables carrying heavy currents are firmly clamped so that they do not pull or push on each other. Link to earlier classes : the field of a straight wire from the previous section.

Simple definition

Two long parallel wires carrying currents in the same direction attract each other, and wires carrying currents in opposite directions repel. This force is used to define the SI unit of current.

Rule 1: Force per unit length — F ÷ l = μ₀ I₁ I₂ ÷ (2π d) (d = distance between the wires). In SI units, μ₀ ÷ 2π = 2 × 10⁻⁷ T m A⁻¹.

  • Easy : I₁ = 10 A, I₂ = 20 A, d = 0.10 m ⇒ F ÷ l = 2 × 10⁻⁷ × 200 ÷ 0.10 = 4 × 10⁻⁴ N m⁻¹.
  • Medium : If d is halved, the force per unit length doubles to 8 × 10⁻⁴ N m⁻¹.
  • Exam : Wires 20 cm apart carrying 5 A and 8 A in the same direction attract with 4 × 10⁻⁵ N m⁻¹, so a 2 m length feels 8 × 10⁻⁵ N.

Rule 2: Definition of the ampere — One ampere is the current which, flowing in each of two infinitely long parallel wires 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length.

  • Easy : Put I₁ = I₂ = 1 A and d = 1 m : F ÷ l = 2 × 10⁻⁷ N m⁻¹.
  • Medium : This fixes μ₀ ÷ 2π = 2 × 10⁻⁷ in SI units.
  • Exam : Opposite currents repel with the same magnitude of force.
Remember

Same direction attract; opposite repel; F/l = μ₀I₁I₂/2πd.

Common mistake

Mixing this up with charges, where like charges repel. For parallel currents, "like" attracts.

Shortcut

Memory clue : "Parallel currents hug; anti-parallel currents push away."

Spot the difference

CurrentsForce
same directionattraction
opposite directionsrepulsion

Practice questions

Easy
  1. 1Do parallel wires with currents in the same direction attract or repel?
  2. 2Write the force per unit length between two long parallel currents.
  3. 3What is the force per unit length for 1 A in each of two wires 1 m apart?
Medium
  1. 4Find the force per metre between wires with 10 A and 20 A, 0.10 m apart.
  2. 5What is the new force per metre if the wires are brought to 0.05 m?
  3. 6State the definition of 1 ampere.
Challenging
  1. 7What is the nature of the force if one current is reversed?
  2. 8Wires 20 cm apart carry 5 A and 8 A in the same direction. Find the force on a 2 m length.
Show answers (try them yourself first)
  1. 1They attract.
  2. 2F ÷ l = μ₀I₁I₂ ÷ 2πd.
  3. 32 × 10⁻⁷ N m⁻¹.
  4. 44 × 10⁻⁴ N m⁻¹.
  5. 58 × 10⁻⁴ N m⁻¹.
  6. 6The current which, in two long parallel wires 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre.
  7. 7It becomes repulsion.
  8. 8F ÷ l = 2 × 10⁻⁷ × 40 ÷ 0.20 = 4 × 10⁻⁵ N m⁻¹, so F = 8 × 10⁻⁵ N.

Torque on a current loop and the magnetic dipole moment

From daily life

An electric motor spins because the coil in its magnetic field feels a turning effect. Link to earlier classes : torque from Class 11 and the dipole in an electric field from Chapter 1.

Simple definition

A current loop in a uniform magnetic field feels no net force but does feel a torque that tends to turn it so that its plane is perpendicular to the field. The loop behaves as a magnetic dipole with magnetic moment m = N I A (unit A m²), pointing along the normal to the loop by the right-hand rule.

Rule 1: Torque on a loop — τ = N I A B sin θ = m B sin θ (θ = angle between m and B). The torque is greatest when the plane of the loop is parallel to the field.

  • Easy : N = 50, I = 2 A, A = 0.04 m², B = 0.30 T ⇒ m = 4.0 A m² and τ(max) = 1.2 N m.
  • Medium : When the angle between m and B is 30°, τ = 4.0 × 0.30 × 0.5 = 0.6 N m.
  • Exam : When the plane of the loop is perpendicular to B, m is parallel to B and τ = 0.

Rule 2: Magnetic moment of a circular loop — m = N I π R². The magnetic energy is U = − m B cos θ, with the same form as for an electric dipole.

  • Easy : A single loop of radius 0.10 m carrying 2 A has m = 2 × π × 0.01 = 0.063 A m².
  • Medium : Doubling the radius makes m four times as large.
  • Exam : The analogy: p ↔ m, E ↔ B, τ = p × E ↔ τ = m × B.
Remember

m = NIA; τ = mB sin θ; no net force in a uniform field.

Common mistake

Taking θ as the angle between the plane of the loop and B. It is the angle between m (the normal) and B.

Shortcut

Memory clue : "A loop is a magnet: moment equals N, I, A."

Spot the difference

Electric dipoleCurrent loop
p = q × 2am = NIA
τ = pE sin θτ = mB sin θ

Practice questions

Easy
  1. 1Write the SI unit of the magnetic moment.
  2. 2What is the net force on a current loop in a uniform field?
  3. 3What is the torque when the plane of the loop is perpendicular to B?
Medium
  1. 4Find the maximum torque on a coil with N = 50, I = 2 A, A = 0.04 m² in a 0.30 T field.
  2. 5Find the magnetic moment of that coil.
  3. 6Find the magnetic moment of a single circular loop of radius 0.10 m carrying 2 A.
Challenging
  1. 7Find the torque on the coil of question 4 when m makes 30° with B.
  2. 8By what factor does m change if the radius of a loop is doubled at the same current?
Show answers (try them yourself first)
  1. 1A m².
  2. 2Zero.
  3. 3Zero.
  4. 4τ = NIAB = 1.2 N m.
  5. 5m = NIA = 4 A m².
  6. 6m = Iπr² ≈ 0.063 A m².
  7. 7τ = mB sin 30° = 0.6 N m.
  8. 8It becomes four times.

The moving coil galvanometer, ammeter and voltmeter

From daily life

The pointer of an analogue ammeter or voltmeter swings because of a coil in a magnetic field. Link to earlier classes : torque on a coil from the previous section and series and parallel resistors from Chapter 3.

Simple definition

A moving coil galvanometer detects small currents. A coil with N turns hangs between curved magnet poles that give a radial field, so the torque is the same at every angle. The magnetic torque N I A B is balanced by the spring torque k φ.

Rule 1: Galvanometer action — N I A B = k φ, so φ ÷ I = N A B ÷ k (φ = deflection, k = torsion constant). A galvanometer of resistance G and full-scale current I_g becomes an ammeter with a small shunt S in parallel, and a voltmeter with a large resistance R in series.

  • Easy : More turns or a stronger magnet increase the current sensitivity.
  • Medium : Doubling N doubles the current sensitivity but also doubles G, so the voltage sensitivity φ ÷ V = NAB ÷ (kR) stays about the same.
  • Exam : The radial field keeps the scale uniform.

Rule 2: Conversion — Shunt S = I_g G ÷ (I − I_g); series resistance R = V ÷ I_g − G.

  • Easy : G = 50 Ω and I_g = 2 mA to read up to 1 A needs S = 0.002 × 50 ÷ 0.998 = 0.100 Ω.
  • Medium : To read up to 10 V, R = 10 ÷ 0.002 − 50 = 4950 Ω.
  • Exam : To read up to 5 V, R = 5 ÷ 0.002 − 50 = 2450 Ω.
Remember

Ammeter : low resistance, in series with the circuit, with a shunt. Voltmeter : high resistance, in parallel with the component.

Common mistake

Connecting an ammeter in parallel, which can burn it.

Shortcut

Memory clue : "Ammeter: Alone in series, tiny shunt. Voltmeter: Visits in parallel, big series R."

Spot the difference

InstrumentAdded resistorConnectedResistance
Ammetershunt (parallel to G)in seriesvery small
Voltmeterseries resistorin parallelvery large

Practice questions

Easy
  1. 1Why is a radial magnetic field used in a galvanometer?
  2. 2How is a shunt connected to a galvanometer?
  3. 3How is a high resistance connected to make a voltmeter?
Medium
  1. 4A galvanometer has G = 50 Ω and I_g = 2 mA. Find the shunt to read up to 1 A.
  2. 5Find the series resistance to read up to 10 V.
  3. 6Find the series resistance to read up to 5 V.
Challenging
  1. 7If the number of turns is doubled, what happens to the current sensitivity and the voltage sensitivity?
  2. 8Should an ideal ammeter have a large or a small resistance?
Show answers (try them yourself first)
  1. 1So that the torque, and the scale, is the same at every deflection.
  2. 2In parallel.
  3. 3In series with the galvanometer.
  4. 4S = 0.002 × 50 ÷ 0.998 ≈ 0.100 Ω.
  5. 5R = 10 ÷ 0.002 − 50 = 4950 Ω.
  6. 6R = 5 ÷ 0.002 − 50 = 2450 Ω.
  7. 7The current sensitivity doubles; the voltage sensitivity stays about the same since the resistance also doubles.
  8. 8Small (ideally zero).