Magnetic fields & induction worksheet with answers

Free printable Magnetic fields & induction worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
+v = 1 × 10⁵ m/sB = 0.5 T

An alpha particle moves at 1 × 10⁵ m/s at an angle of 60° to a uniform magnetic field of 0.5 T. Find the magnitude of the magnetic force F on it.

|q| = 3.204 × 10⁻¹⁹ Cv = 1 × 10⁵ m/sB = 0.5 Tθ = 60°
F = fN
2.Easy
nB60°20 cm × 15 cmB = 0.4 T

A rectangular loop of 20 cm by 15 cm lies in a uniform magnetic field of 0.4 T. The field makes an angle of 60° with the normal of the loop. Find the magnetic flux Φ through the loop.

B = 0.4 TA = 20 cm × 15 cm60° to the normal
Φ = mWb
3.Medium
nBA = 50 cm², N = 200B: 0.05 → 0.65 T

A coil of 200 turns, area 50 cm² and resistance 2 Ω forms a closed circuit. A uniform field perpendicular to the coil rises steadily from 0.05 T to 0.65 T in 50 ms. Find the average induced current I.

N = 200A = 50 cm²B: 0.05 T → 0.65 TΔt = 50 msR = 2 Ω
I = A
4.Medium
nBA = 20 cm², N = 50B = 0.1 T

A coil of 50 turns has an area of 20 cm². The coil sits in a uniform field of 0.1 T perpendicular to its plane, and the field is switched off in 40 ms. Find the average induced emf ε.

N = 50A = 20 cm²B = 0.1 TΔt = 40 ms
ε = mV
5.Hard
Bn = 1500 rpm20 cm × 5 cmN = 100B = 0.25 T

A coil of 100 turns, 20 cm × 5 cm, rotates at 1500 rpm in a uniform field of 0.25 T about an axis perpendicular to the field. Find the rms value Erms of the emf.

N = 100A = 20 cm × 5 cmB = 0.25 Tn = 1500 rpm
Erms = V
6.Hard
l = 20 cmN = 1500, ⌀ 2 cm

A long air-core solenoid has 1500 turns, a length of 20 cm and a diameter of 2 cm. Find its self-inductance L.

N = 1500l = 20 cm⌀ = 2 cmμ₀ = 4π × 10⁻⁷ T·m/A
L = mH

Answer key

  1. 1. F = 13.87 fN
  2. 2. Φ = 6 mWb
  3. 3. I = 6 A
  4. 4. ε = 250 mV
  5. 5. Erms = 27.77 V
  6. 6. L = 4.441 mH

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