Magnetic fields & induction: practice problems with step-by-step solutions
Also called: electromagnetic induction · Faraday’s law · Lorentz force
Free Magnetic fields & induction practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: fields of currents and magnetic forces, charged particles, Faraday and Lenz, generators, inductance and particle beams.
Free practice problemEasy
A flat circular coil of 100 turns and radius 2 cm carries a current of 0.5 A. Find the magnetic flux density B at its centre.
Every electric current is surrounded by a magnetic field B, measured in tesla (T). The Earth’s field is about 50 µT; a fridge magnet a few mT.
Around a long straight wire the field lines are circles. Point the thumb of your right hand along the current: your fingers curl in the direction of B (the right-hand grip rule). In drawings ⊗ means a field into the page and ⊙ out of it.
wire: B = μ₀I/(2πr) loop centre: B = μ₀NI/(2R) solenoid: B = μ₀(N/L)·I
μ₀ = 4π × 10⁻⁷ T·m/A. Always use metres: 5 cm is 0.05 m. Inside a long solenoid the field is uniform and does not depend on the radius.
Self-check The distance from a long straight wire doubles. What happens to B?
It doubles
It falls to a quarter
It halves
It stays the same
Answer: It halves. B = μ₀I/(2πr) falls with 1/r, so twice the distance gives half the field. The 1/r² law belongs to point charges, not to wires.
2Magnetic forces
A field pushes on moving charges. On a straight wire of length L carrying I, and on a single charge q moving at v:
F = BIL·sin θ F = |q|vB·sin θ
θ is the angle between the current (or velocity) and the field: a wire parallel to B feels no force. The force is perpendicular to both. Fingers of the right hand along v, curl them towards B: the thumb gives F for a positive charge. For an electron, reverse it.
Because the force is always perpendicular to the velocity, a charge moving across a uniform field goes round a circle. Setting |q|vB = mv²/r gives r = mv/(|q|B). The time for one turn, T = 2πm/(|q|B), does not depend on the speed.
Two parallel wires attract when their currents flow the same way and repel when they flow in opposite directions, with F/L = μ₀I₁I₂/(2πd).
Self-check A proton and an electron move at the same speed across the same field. Which moves on the larger circle?
Both circles are the same size
Neither: they move in straight lines
The electron
The proton
Answer: The proton. r = mv/(|q|B). Both have the same charge size, but the proton is about 1836 times heavier, so its circle is 1836 times larger. They also bend in opposite directions.
3Flux and Faraday’s law
The magnetic flux through a loop counts how much field passes through it: Φ = BA·cos θ, with θ between B and the normal of the loop. Unit: weber (Wb = T·m²).
When the flux through a coil changes, an emf is induced (Faraday’s law). The induced current flows so that its own field opposes the change (Lenz’s law):
ε = −N·ΔΦ/Δt
The flux can change because B changes, because the area changes, or because the coil turns. Turning a coil over in a field reverses the flux, so ΔΦ = 2BA.
A rod of length L moving at v across a field sweeps out area, so ε = BLv. With a resistor R the current is I = ε/R, and the field pushes back on that current with F = BIL: you must keep pulling to keep the rod moving.
Self-check The field into the page through a loop is increasing. Which way does the induced current flow, seen from the front?
Clockwise
Counterclockwise
It depends on the size of the loop
There is no current
Answer: Counterclockwise. The induced current opposes the increase of flux into the page, so its own field inside the loop points out of the page. By the grip rule that needs a counterclockwise current.
4Generators and inductance
A coil of N turns and area A turning at angular speed ω in a field B gives a sinusoidal emf with peak E₀ = NBAω. The rms value is E₀/√2, and ω = 2π·n/60 for n in rpm.
A changing current in a coil changes its own flux, which induces an emf that opposes the change: this is self-inductance L, in henry (H).
ε = L·ΔI/Δt long solenoid: L = μ₀N²A/l energy: W = ½LI²
Two coils that share flux have a mutual inductance M: a changing current in one induces ε₂ = M·di₁/dt in the other, with M = k√(L₁L₂) and 0 ≤ k ≤ 1. Transformers use this.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A metal rod 40 cm long slides at 5 m/s on rails in a field of 0.5 T. The rails are joined by a 2 Ω resistor. Find the current and the force needed to keep the rod moving at constant speed.
1
Induced emf
ε = BLv = 0.5 · 0.40 · 5 = 1.0 V
2
Current
I = ε/R = 1.0/2 = 0.5 A
3
Braking force
The current in the rod sits in the field, so the rod feels F = BIL, opposite to its motion (Lenz).
F = 0.5 · 0.5 · 0.40 = 0.10 N
4
Check with energy
The pulling power F·v = 0.10 · 5 = 0.5 W equals the heat in the resistor, I²R = 0.25 · 2 = 0.5 W. Energy is conserved.
Answer
I = 0.5 A and F = 0.10 N.
6Common mistakes
Leaving distances in cm or areas in cm². 1 cm² = 10⁻⁴ m².
Mixing up the formulas: μ₀I/(2πr) for a wire, μ₀NI/(2R) for the centre of a loop, μ₀nI for a solenoid.
Using the angle to the plane in Φ = BA·cos θ instead of the angle to the normal.
Using the final field instead of the change of field in Faraday’s law.
Forgetting to reverse the force direction for a negative charge.
Mixing peak and rms values for a generator.
7How exam questions look
Easy: B of a wire, a loop or a solenoid, the force on a wire or a moving charge, and the flux through a loop.
Medium: the radius and period of a charge in a field, Faraday’s law for a coil, a rod on rails, parallel wires, and the direction of a force or an induced current.
Hard: generators, self-inductance and stored energy, mutual inductance, mass spectrometers and velocity selectors.
For directions, sketch the field symbols and use the right hand every time; for sizes, write the formula first and convert every quantity to SI units.
A long straight wire carries 8 A. At what distance r from the wire is the magnetic flux density 32 µT?
I = 8 AB = 32 µTμ₀ = 4π × 10⁻⁷ T·m/A
Find r(cm)
Show hints
Rearrange B = μ₀I/(2πr) for r.
r = μ₀I/(2πB).
Convert the result from metres to centimetres.
Show the answer and the step-by-step solution
Magnetic field of currents: around a long straight wire the field lines are circles (right-hand grip rule) with B = μ₀I/(2πr). At the centre of a flat circular coil of N turns, B = μ₀NI/(2R). Inside a long solenoid the field is uniform, B = μ₀nI with n = N/L turns per metre, whatever the radius. μ₀ = 4π × 10⁻⁷ T·m/A. Work in SI units (metres and amperes), and add the fields of several sources as vectors.
Distance from the field
r = μ₀I/(2πB) = 4π × 10⁻⁷·8/(2π·32 × 10⁻⁶) = 0.05 m = 5 cm
Answer
r = 5 cm
Example 2Medium
A coil of 50 turns has an area of 80 cm². A uniform field perpendicular to the coil rises steadily from 0.2 T to 1 T in 100 ms. Find the average induced emf ε.
N = 50A = 80 cm²B: 0.2 T → 1 TΔt = 100 ms
Find ε(V)
Show hints
Faraday's law: ε = N·ΔΦ/Δt.
ΔΦ = ΔB·A.
A = 0.008 m² and Δt = 0.1 s.
Show the answer and the step-by-step solution
Electromagnetic induction: the flux through a loop is Φ = BA·cos θ, with θ between B and the normal of the loop. Faraday's law: a change of flux induces emf = −N·dΦ/dt in a coil of N turns, and Lenz's law says the induced current flows so that its own field opposes the change. A rod moving across a field gives emf = BLv. A coil rotating at ω gives e = NBAω·sin ωt, peak NBAω. Self-inductance: emf = −L·di/dt, with L = μ₀N²A/l for a long solenoid, and the energy stored is W = ½LI². Two coupled coils: emf₂ = −M·di₁/dt with M = k√(L₁L₂).
Change of flux per turn
ΔΦ = (1 − 0.2)·0.008 = 0.0064 Wb
Average emf
ε = NΔΦ/Δt = 50·0.0064/0.1 = 3.2 V
Answer
ε = 3.2 V
Example 3Hard
Two coils with self-inductances of 50 mH and 50 mH have a coupling coefficient of 0.3. The current in the first coil changes at 1000 A/s. Find the emf ε induced in the second coil.
L1 = 50 mHL2 = 50 mHk = 0.3di/dt = 1000 A/s
Find ε(V)
Show hints
M = k√(L₁L₂).
ε₂ = M·di₁/dt.
Convert M to henry.
Show the answer and the step-by-step solution
Electromagnetic induction: the flux through a loop is Φ = BA·cos θ, with θ between B and the normal of the loop. Faraday's law: a change of flux induces emf = −N·dΦ/dt in a coil of N turns, and Lenz's law says the induced current flows so that its own field opposes the change. A rod moving across a field gives emf = BLv. A coil rotating at ω gives e = NBAω·sin ωt, peak NBAω. Self-inductance: emf = −L·di/dt, with L = μ₀N²A/l for a long solenoid, and the energy stored is W = ½LI². Two coupled coils: emf₂ = −M·di₁/dt with M = k√(L₁L₂).
Mutual inductance
M = 0.3·√(50·50) = 15 mH
Induced emf
ε = M·di/dt = 0.015·1000 = 15 V
Answer
ε = 15 V
Questions and answers
What is Magnetic fields & induction?
Every electric current is surrounded by a magnetic field B, measured in tesla (T). The Earth’s field is about 50 µT; a fridge magnet a few mT.
Around a long straight wire the field lines are circles. Point the thumb of your right hand along the current: your fingers curl in the direction of B (the right-hand grip rule). In drawings ⊗ means a field into the page and ⊙ out of it.
How do I solve Magnetic fields & induction problems step by step?
Convert every quantity to SI units: cm to m, cm² to m², mT and µT to T.
Pick the right formula: B = μ₀I/(2πr) for a wire, μ₀NI/(2R) for the centre of a coil, μ₀nI inside a solenoid; F = BIL·sin θ or |q|vB·sin θ for forces.
For induction, find the change of flux ΔΦ = Δ(BA·cos θ) and use ε = N·ΔΦ/Δt, or ε = BLv for a moving rod and NBAω for a rotating coil.
Find directions separately with the right-hand rules, and Lenz’s law for induced currents.
What are common mistakes in Magnetic fields & induction?
Leaving lengths in cm or areas in cm².
Using the wire formula for a loop or a solenoid.
Using the angle to the plane in Φ = BA·cos θ, which needs the angle to the normal.
Forgetting to reverse the force direction for a negative charge.
Mixing up peak and rms values of a generator.
What do the Magnetic fields & induction exercises cover?
Easy: fields of currents and magnetic forces
Medium: charged particles, Faraday and Lenz
Hard: generators, inductance and particle beams
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.