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Moving Charges and Magnetism – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 4. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Moving charges produce magnetic fields (Biot–Savart, Ampere) and experience forces in magnetic fields (Lorentz force).

Fields of currents

  • Biot–Savart law: dB = (µ₀/4π) I dl sin θ / r².
  • Long straight wire B = µ₀I/(2πr); centre of a circular loop B = µ₀I/(2R).
  • Solenoid B = µ₀nI (n = turns per metre); Ampere's law ∮B·dl = µ₀I.

Forces and devices

  • Force on a charge F = qvB sin θ; circular motion radius r = mv/(qB).
  • Force on a wire F = BIl sin θ; parallel currents attract, antiparallel repel.
  • Moving coil galvanometer: torque NIAB; converted to ammeter (low shunt in parallel) or voltmeter (high resistance in series).

Key terms

Lorentz force
force on a charge moving in electric and magnetic fields
Shunt
small resistance connected in parallel to a galvanometer
Biot–Savart law
gives the magnetic field dB = (µ₀/4π) I dl × r̂ / r² due to a small current element

Formulas

  • B = µ₀I/(2πr)
  • B = µ₀nI
  • F = qvB sin θ
  • r = mv/(qB)

Important questions with answers

Why does a magnetic force do no work on a moving charge?2 marks

The force qv × B is always perpendicular to velocity, so it changes only the direction of motion, not the speed or kinetic energy.

Using Ampere's circuital law, find the magnetic field inside a long solenoid.3 marks

Take a rectangular loop with one side of length l inside, parallel to the axis. Field outside is negligible and the perpendicular sides contribute nothing. ∮B·dl = Bl = µ₀(nl)I, where n is turns per unit length. So B = µ₀nI, uniform inside.

Derive the force per unit length between two long parallel current-carrying wires and use it to define the ampere.5 marks

Wire 1 with current I₁ produces at distance d a field B₁ = µ₀I₁/(2πd). Wire 2 of length L with current I₂ experiences F = I₂LB₁ = µ₀I₁I₂L/(2πd). Force per unit length f = µ₀I₁I₂/(2πd); it is attractive for currents in the same direction and repulsive for opposite currents. One ampere is the current which, flowing in two infinitely long parallel straight wires 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre of length (the SI now fixes it via the elementary charge, but this remains its physical meaning).

Quick revision

  • Parallel currents attract.
  • µ₀ = 4π × 10⁻⁷ T m A⁻¹.
  • Ammeter: shunt in parallel; voltmeter: series resistance.

Exam tips

  • A magnetic force does no work on a moving charge (force ⟂ velocity).
  • A moving charge in a uniform field moves in a circle if v ⟂ B and in a helix if v makes an angle with B.

Practice MCQs

  1. SI unit of magnetic field is

    1. weber
    2. tesla
    3. henry
    4. gauss only
    Show answer

    Answer: (b) tesla. Tesla.

  2. Two parallel wires carrying current in the same direction

    1. repel
    2. do nothing
    3. attract
    4. rotate
    Show answer

    Answer: (c) attract. Attract.

  3. A charge moving parallel to a magnetic field experiences

    1. maximum force
    2. zero force
    3. force along the field
    4. force opposite to motion
    Show answer

    Answer: (b) zero force. sin 0° = 0.

  4. A galvanometer is converted into an ammeter by connecting

    1. high resistance in series
    2. high resistance in parallel
    3. low resistance in parallel
    4. low resistance in series
    Show answer

    Answer: (c) low resistance in parallel. Shunt.

  5. Magnetic field at the centre of a circular loop of radius r carrying current I is

    1. µ₀I/2πr
    2. µ₀I/2r
    3. µ₀I/4πr
    4. µ₀Ir
    Show answer

    Answer: (b) µ₀I/2r. Standard result.

  6. Field at the centre of a circular loop of radius R carrying I is

    1. µ₀I/(2πR)
    2. µ₀IR/2
    3. µ₀I/R
    4. µ₀I/(2R)
    Show answer

    Answer: (d) µ₀I/(2R). Standard result.

  7. Work done by the magnetic force on a moving charge is

    1. qvB
    2. qvBr
    3. zero
    4. negative
    Show answer

    Answer: (c) zero. Force is perpendicular to velocity.

  8. A galvanometer of resistance 50 Ω gives full deflection at 1 mA. Series resistance needed to convert it into a 10 V voltmeter is

    1. 950 Ω
    2. 10,000 Ω
    3. 9950 Ω
    4. 50 Ω
    Show answer

    Answer: (c) 9950 Ω. R = V/Ig − G.

All chapters of Class 12 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.