A-Level — physicsHard3 min read

Magnetic Fields and Electromagnetic Induction

F = BIL; F = Bqv; Faraday's law; Lenz's law; transformers and generators

Tutor AI Team
Tutor AI Team

# Magnetic Fields and Electromagnetic Induction — A-Level Physics

Magnetic fields and electromagnetic induction are the physics behind motors, generators, transformers, and indeed most of modern technology.

1. Magnetic Flux Density ($B$)

Magnetic flux density (or magnetic field strength) is measured in tesla (T).

Force on a Current-Carrying Wire

F=BILsin⁡θ\boxed{F = BIL\sin\theta}

Where θ\theta is the angle between the wire and the field. Maximum when perpendicular (θ=90°\theta = 90°).

Direction: Fleming's Left-Hand Rule (First finger: Field, seCond finger: Current, thuMb: Motion/Force).

Force on a Moving Charge

F=Bqvsin⁡θ\boxed{F = Bqv\sin\theta}

This is the Lorentz force. For a charged particle moving perpendicular to B: F=BqvF = Bqv.

The force is always perpendicular to velocity → circular motion: Bqv=mv2/r  ⟹  r=mvBqBqv = mv^2/r \implies r = \frac{mv}{Bq}

2. Magnetic Flux

Φ=BAcos⁡θ\Phi = BA\cos\theta

Units: weber (Wb) = T·m²

Flux linkage for a coil of NN turns: NΦ=BANcos⁡θN\Phi = BAN\cos\theta

3. Faraday's Law

ε=−d(NΦ)dt\boxed{\varepsilon = -\frac{d(N\Phi)}{dt}}

The induced EMF equals the rate of change of flux linkage.

4. Lenz's Law

The induced current flows in a direction that opposes the change that caused it. (This is the minus sign in Faraday's law.)

Consequence of conservation of energy.

5. Generators and Alternators

A coil rotating in a magnetic field: NΦ=BANcos⁡(ωt)N\Phi = BAN\cos(\omega t)

ε=BANωsin⁡(ωt)\varepsilon = BAN\omega\sin(\omega t)

Peak EMF: ε0=BANω\varepsilon_0 = BAN\omega

This produces alternating current (AC).

6. Transformers

VsVp=NsNp=IpIs\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

(For an ideal transformer, VpIp=VsIsV_p I_p = V_s I_s)

Step-up: Ns>NpN_s > N_p → increases voltage Step-down: Ns<NpN_s < N_p → decreases voltage

Energy Losses

  • Eddy currents in core → heat (reduced by laminated core)
  • Resistance of coils → heat (use thick, low-resistance wire)
  • Magnetic flux leakage (use soft iron core to concentrate flux)
  • Hysteresis losses (use soft magnetic material)

Worked Example: Force on Wire

Problem

A 50 cm wire carries 3 A at 90° to a 0.2 T field. Find the force.

F=BIL=0.2×3×0.5=0.3F = BIL = 0.2 \times 3 \times 0.5 = 0.3 N

Solution

Worked Example: Charged Particle

Problem

An electron (v=5×106v = 5 \times 10^6 m/s) enters a 0.01 T field perpendicular. Find the radius of its circular path.

r=mv/(Bq)=9.11×10−31×5×106/(0.01×1.6×10−19)r = mv/(Bq) = 9.11 \times 10^{-31} \times 5 \times 10^6/(0.01 \times 1.6 \times 10^{-19}) =4.555×10−24/1.6×10−21=2.85×10−3= 4.555 \times 10^{-24}/1.6 \times 10^{-21} = 2.85 \times 10^{-3} m = 2.85 mm

Solution

Worked Example: Faraday's Law

Problem

A coil of 200 turns and area 0.05 m² is in a field that changes from 0.3 T to 0.1 T in 0.02 s. Find the induced EMF.

ε=N×ΔΦ/Δt=200×(0.3−0.1)×0.05/0.02=200×0.01/0.02=100\varepsilon = N \times \Delta\Phi/\Delta t = 200 \times (0.3 - 0.1) \times 0.05/0.02 = 200 \times 0.01/0.02 = 100 V

Solution

8. Practice Questions

    1. A 0.4 m wire carrying 5 A is at 30° to a 0.15 T field. Calculate the force. (2 marks)
    1. A proton (m=1.67×10−27m = 1.67 \times 10^{-27} kg) moves at 3×1073 \times 10^7 m/s in a 0.5 T field. Find the radius of curvature. (2 marks)
    1. A transformer has 500 primary turns and 50 secondary turns. The primary voltage is 230 V. Find the secondary voltage. (2 marks)

    Answers

    1. F=BILsin⁡θ=0.15×5×0.4×sin⁡30°=0.15F = BIL\sin\theta = 0.15 \times 5 \times 0.4 \times \sin30° = 0.15 N.
    1. r=mv/(Bq)=1.67×10−27×3×107/(0.5×1.6×10−19)=5.01×10−20/8×10−20=0.626r = mv/(Bq) = 1.67 \times 10^{-27} \times 3 \times 10^7/(0.5 \times 1.6 \times 10^{-19}) = 5.01 \times 10^{-20}/8 \times 10^{-20} = 0.626 m.

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Summary

  • Force on wire: F=BILsin⁡θF = BIL\sin\theta; on charge: F=Bqvsin⁡θF = Bqv\sin\theta
  • Fleming's LHR for force direction
  • Faraday: ε=−d(NΦ)/dt\varepsilon = -d(N\Phi)/dt; Lenz: opposes change
  • Generator: ε=BANωsin⁡(ωt)\varepsilon = BAN\omega\sin(\omega t)
  • Transformer: Vs/Vp=Ns/NpV_s/V_p = N_s/N_p
Tags:
#a-level#physics#fields#magnetic-field#electromagnetic-induction#faraday#lenz

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