JEE Main Physics · Chapter 15

Magnetic Effects of Current and Magnetism

Biot–Savart and Ampère’s laws, Lorentz force, galvanometer conversions and magnetic materials.

NCERT Class 12JEE Main: 1–2 Qs per shift 9 min read

1. Chapter at a Glance

Moving charges produce magnetic fields, and magnetic fields push on moving charges. This chapter covers the Biot–Savart law, Ampère's circuital law, fields of wires, loops and solenoids, the Lorentz force, motion of charges in magnetic fields (cyclotron), forces and torques on current-carrying conductors, the moving-coil galvanometer and its conversion to ammeter/voltmeter, and the basics of magnetism and magnetic materials. JEE Main asks two questions on average.

What JEE asks most
  • Magnetic field at the centre of loops and arcs, near straight wires and inside solenoids.
  • Radius, period and pitch of a charged particle's path in a magnetic field.
  • Force between parallel currents; torque on a current loop.
  • Galvanometer conversions; dia-, para- and ferromagnetic materials.

2. Magnetic Field of Currents

Biot–Savart: dB = (μ₀/4π) I dl × r̂ /r²    Ampère: ∮B·dl = μ₀Ienclosed
SourceMagnetic field
Long straight wire (distance r)B = μ₀I/2πr
Finite wireB = (μ₀I/4πr)(sinθ₁ + sinθ₂)
Centre of circular loop (N turns, radius R)B = μ₀NI/2R
Axis of loop (distance x)B = μ₀NIR²/2(R² + x²)3/2
Arc of angle θ at centreB = (μ₀I/4πR) θ
Long solenoid (n turns per metre)B = μ₀nI inside; μ₀nI/2 at an end
ToroidB = μ₀NI/2πr

Direction: right-hand thumb rule — thumb along current, curled fingers give the field. μ₀ = 4π × 10⁻⁷ T m A⁻¹.

3. Force on Moving Charges and Currents

  • Lorentz force: F = q(E + v × B). The magnetic part qvB sinθ is perpendicular to v, so it does no work and cannot change speed.
  • Charge moving ⊥ B: circle of radius r = mv/qB; period T = 2πm/qB (independent of speed); cyclotron frequency f = qB/2πm.
  • Velocity at angle θ to B: helix with radius mv sinθ/qB and pitch (v cosθ)T.
  • Velocity selector: crossed E and B let through only particles with v = E/B.
  • Force on a wire: F = I L × B (BIL sinθ). Parallel wires: F/L = μ₀I₁I₂/2πd — attractive for currents in the same direction, repulsive for opposite.
  • Torque on a current loop: τ = m × B, |τ| = NIAB sinθ, magnetic moment m = NIA.
×××××××× +q v F = qvB r = mv/qB B into page (×)
The magnetic force is always perpendicular to the velocity and provides the centripetal force: qvB = mv²/r.

4. Moving-Coil Galvanometer

  • Deflection θ ∝ current: NIAB = kθ (radial field makes the torque independent of angle). Current sensitivity θ/I = NAB/k.
  • Ammeter: a small shunt S in parallel: S = IgG/(I − Ig). An ideal ammeter has zero resistance.
  • Voltmeter: a large resistance R in series: R = V/Ig − G. An ideal voltmeter has infinite resistance.

5. Magnetism and Matter

  • A current loop behaves like a magnetic dipole; a bar magnet is equivalent to a solenoid. Field of a short magnet: axial 2(μ₀/4π)m/r³, equatorial (μ₀/4π)m/r³.
  • Magnetic intensity H, magnetisation M: B = μ₀(H + M); susceptibility χ = M/H; relative permeability μr = 1 + χ.
  • Diamagnetic (Cu, Bi, water): χ small and negative; weakly repelled; independent of temperature. Superconductors are perfect diamagnets.
  • Paramagnetic (Al, O₂): χ small and positive; χ ∝ 1/T (Curie's law).
  • Ferromagnetic (Fe, Co, Ni): χ very large; show hysteresis; become paramagnetic above the Curie temperature. Soft iron (low retentivity and coercivity) suits electromagnets; steel suits permanent magnets.

6. Formula Master Table

QuantityFormula
Straight wireB = μ₀I/2πr
Loop centreB = μ₀NI/2R
SolenoidB = μ₀nI
Lorentz forceF = qvB sinθ
Radius, periodr = mv/qB; T = 2πm/qB
Parallel wiresF/L = μ₀I₁I₂/2πd
Torque on loopτ = NIAB sinθ
Shunt / series resistorS = IgG/(I − Ig); R = V/Ig − G

7. Mind Map

Magnetic Effects
Sources
  • Biot–Savart
  • Ampère's law
  • Wire, loop, solenoid
Forces
  • Lorentz force
  • Circular, helical paths
  • Parallel wires
Devices
  • Galvanometer
  • Ammeter, voltmeter
  • Cyclotron
Materials
  • Dia, para, ferro
  • Hysteresis
  • Curie law

8. Tips and Tricks

Shortcuts
  • For arcs, B = (θ/2π) × μ₀I/2R — a semicircle gives half of a full loop.
  • A straight wire through the centre of a loop's plane (radially) gives zero field at the centre.
  • r = p/qB = √(2mK)/qB: for equal kinetic energies, r ∝ √m/q.
  • Magnetic force never changes kinetic energy.
Mistakes students make
  • Forgetting sinθ: a charge moving parallel to B feels no force.
  • Getting the direction wrong for negative charges (reverse the right-hand result).
  • Connecting the shunt in series or the multiplier in parallel.

9. Important One-liners

  • A magnetic field does no work on a moving charge.
  • Cyclotron frequency does not depend on the particle's speed.
  • Parallel currents attract; antiparallel currents repel.
  • The field inside a long solenoid is uniform.
  • Diamagnetic substances have negative susceptibility.
  • Ferromagnetism disappears above the Curie temperature.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Average level

N1Average

A circular coil of 100 turns and radius 10 cm carries 2 A. Find the magnetic field at its centre.

Show answer & solution

Concept: B = μ₀NI/2R.

  1. B = 4π × 10⁻⁷ × 100 × 2/(2 × 0.1).
  2. = 4π × 10⁻⁴ ≈ 1.26 × 10⁻³ T.

Answer: about 1.26 × 10⁻³ T

N2Average

Find the magnetic field 5 cm from a long straight wire carrying 10 A.

Show answer & solution

Concept: B = μ₀I/2πr = 2 × 10⁻⁷ I/r.

  1. B = 2 × 10⁻⁷ × 10/0.05.
  2. = 4 × 10⁻⁵ T.

Answer: 4 × 10⁻⁵ T

N3Average

A long solenoid has 1000 turns per metre and carries 2 A. Find the field inside it.

Show answer & solution

Concept: B = μ₀nI.

  1. B = 4π × 10⁻⁷ × 1000 × 2.
  2. ≈ 2.51 × 10⁻³ T.

Answer: about 2.5 × 10⁻³ T

N4Average

A 0.5 m wire carrying 4 A lies perpendicular to a 0.2 T field. Find the force on it.

Show answer & solution

Concept: F = BIL sinθ.

  1. F = 0.2 × 4 × 0.5 × sin90°.
  2. = 0.4 N.

Answer: 0.4 N

N5Average

A 50-turn coil of area 0.01 m² carries 2 A in a 0.5 T field, with the field parallel to its plane. Find its magnetic moment and the torque on it.

Show answer & solution

Concept: m = NIA; τ = mB sinθ, with θ = 90° between the normal and B.

  1. m = 50 × 2 × 0.01 = 1 A m².
  2. τ = 1 × 0.5 × 1 = 0.5 N m (maximum).

Answer: 1 A m²; 0.5 N m

Advanced level

N6Advanced

A proton moving at 10⁶ m s⁻¹ enters a 0.5 T field at right angles. Find the radius and period of its circular path (m = 1.67 × 10⁻²⁷ kg).

Show answer & solution

Concept: r = mv/qB; T = 2πm/qB.

  1. r = 1.67 × 10⁻²⁷ × 10⁶/(1.6 × 10⁻¹⁹ × 0.5) ≈ 2.09 × 10⁻² m.
  2. T = 2π × 1.67 × 10⁻²⁷/(0.8 × 10⁻¹⁹) ≈ 1.31 × 10⁻⁷ s.

Answer: about 2.1 cm; about 1.3 × 10⁻⁷ s

N7Advanced

Two long parallel wires 10 cm apart carry 10 A and 20 A in the same direction. Find the force per metre between them, and state its nature.

Show answer & solution

Concept: F/L = μ₀I₁I₂/2πd = 2 × 10⁻⁷ I₁I₂/d.

  1. F/L = 2 × 10⁻⁷ × 10 × 20/0.1.
  2. = 4 × 10⁻⁴ N m⁻¹.
  3. Same direction ⇒ attractive.

Answer: 4 × 10⁻⁴ N m⁻¹, attractive

N8Advanced

A galvanometer of resistance 50 Ω gives full deflection at 5 mA. How can it be converted into (a) an ammeter of range 0–5 A, (b) a voltmeter of range 0–10 V?

Show answer & solution

Concept: shunt in parallel for an ammeter; high resistance in series for a voltmeter.

  1. (a) S = IgG/(I − Ig) = 0.005 × 50/4.995.
  2. S ≈ 0.050 Ω in parallel.
  3. (b) R = V/Ig − G = 10/0.005 − 50 = 2000 − 50.
  4. R = 1950 Ω in series.

Answer: 0.05 Ω shunt; 1950 Ω in series

N9Advanced

An electron moving at 2 × 10⁶ m s⁻¹ enters a 1 mT field with its velocity at 30° to the field. Find the radius and pitch of its helical path (me = 9.1 × 10⁻³¹ kg).

Show answer & solution

Concept: v⊥ = v sinθ sets the circle; v∥ = v cosθ carries it along B.

  1. v⊥ = 10⁶ m s⁻¹; v∥ = 1.73 × 10⁶ m s⁻¹.
  2. r = mv⊥/eB = 9.1 × 10⁻³¹ × 10⁶/(1.6 × 10⁻²²) ≈ 5.7 × 10⁻³ m.
  3. T = 2πm/eB ≈ 3.57 × 10⁻⁸ s.
  4. Pitch = v∥T ≈ 1.73 × 10⁶ × 3.57 × 10⁻⁸ ≈ 6.2 × 10⁻² m.

Answer: about 5.7 mm; about 6.2 cm

N10Advanced

(a) In a velocity selector, E = 10⁴ V m⁻¹ and B = 0.02 T. Which speed passes undeflected? (b) Find the cyclotron frequency of a proton in a 1 T field.

Show answer & solution

Concept: (a) qE = qvB; (b) f = qB/2πm.

  1. (a) v = E/B = 10⁴/0.02 = 5 × 10⁵ m s⁻¹ (independent of charge and mass).
  2. (b) f = 1.6 × 10⁻¹⁹ × 1/(2π × 1.67 × 10⁻²⁷).
  3. f ≈ 1.52 × 10⁷ Hz ≈ 15.2 MHz.

Answer: 5 × 10⁵ m s⁻¹; about 15.2 MHz

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

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

  1. Maximum
  2. qvB
  3. Zero
  4. qvB/2
Show solution

F = qvB sinθ with θ = 0 gives zero.

Answer: (C)

Q2Medium

The frequency 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. Depends on radius only
Show solution

f = qB/2πm — faster particles move in bigger circles, taking the same time.

Answer: (C)

Q3Numerical

The current in a circular loop is doubled and its radius is halved. By what factor does the field at the centre change?

Show solution

B ∝ I/R = 2/(1/2) = 4.

Answer: 4

Q4Easy

The magnetic susceptibility of a diamagnetic substance is:

  1. Large and positive
  2. Small and positive
  3. Small and negative
  4. Zero
Show solution

Diamagnets are weakly magnetised opposite to the field: χ small and negative.

Answer: (C)

Q5Hard

A proton and an alpha particle enter the same uniform magnetic field with the same kinetic energy, perpendicular to it. The ratio of their radii rp : rα is:

  1. 1 : 1
  2. 1 : 2
  3. 2 : 1
  4. 1 : 4
Show solution

r = √(2mK)/qB ∝ √m/q. Proton: √1/1 = 1; alpha: √4/2 = 1. Ratio 1 : 1.

Answer: (A)

Q6Numerical

Two long parallel wires 1 m apart each carry 1 A. Find the force per metre between them in units of 10⁻⁷ N m⁻¹.

Show solution

F/L = 2 × 10⁻⁷ × 1 × 1/1 = 2 × 10⁻⁷ N m⁻¹.

Answer: 2

12. Quick Revision Summary

Remember before the exam
  • Wire μ₀I/2πr; loop μ₀NI/2R; solenoid μ₀nI.
  • F = qvB sinθ does no work; r = mv/qB; T = 2πm/qB.
  • Parallel wires: μ₀I₁I₂/2πd; same direction attract.
  • Loop torque NIAB sinθ; ammeter shunt, voltmeter series resistance.
  • Dia χ < 0; para χ ∝ 1/T; ferro: hysteresis, Curie point.
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