Faraday's Law and Lenz's Law: Quick Revision

A brief note on Faraday's Law and Lenz's Law. Read it before a test. Try each question at the end before you open its answer.

Flux and Faraday's law

• Flux: $\Phi = BA\cos\theta$, in weber (Wb). The angle $\theta$ is between the field and the normal to the loop. $1\ \text{Wb} = 1\ \text{T m}^2 = 1\ \text{V s}$.
• Flux linkage: $N\Phi$ for a coil of $N$ turns.
• Ways to change the flux: change $B$, change $A$, or change $\theta$.
• Faraday's law: $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$. Only a change of flux gives an emf. A steady flux gives none, however large it is.
• Current: $I = \dfrac{\varepsilon}{R}$ in a closed circuit.
• Charge: $Q = \dfrac{N\Delta\Phi}{R}$. It does not depend on the time that the change takes.

Lenz's law

• Statement: the induced current has the direction whose own field opposes the change of flux.
• Flux growing: the induced field is against the original field.
• Flux falling: the induced field is along the original field.
• Direction of the current: curl the fingers of your right hand so that the thumb points along the induced field. An induced field out of the page needs an anticlockwise current.
• Energy: the work done against the opposing force becomes electrical energy and then heat. Without Lenz's law, energy would appear from nothing.
• Open coil: an emf, but no current and no opposing force.

Motional emf

• A rod: $\varepsilon = Blv$, when $B$, $l$ and $v$ are at right angles to one another.
• On rails with resistance $R$: $I = \dfrac{Blv}{R}$ and $F = BIl = \dfrac{B^2l^2v}{R}$, opposing the motion.
• Power: $P = Fv = \dfrac{B^2l^2v^2}{R} = I^2R$. All the work becomes heat.
• A tilted field: use only the part of $B$ at right angles to the rod and to its velocity.
• A rod that turns about one end: $\varepsilon = \tfrac{1}{2}B\omega l^2$.

A worked example

A coil of $40$ turns has $15\ \text{cm}^2$ in each turn. It faces a field that falls from $0.50$ T to $0.20$ T in $0.30$ s. The resistance of the circuit is $6.0\ \Omega$. Find the emf, the current and the charge.

1. Change of flux in one turn. $\Delta\Phi = A\,\Delta B = 1.5 \times 10^{-3}\ \text{m}^2 \times 0.30\ \text{T} = 4.5 \times 10^{-4}\ \text{Wb}$.
2. Emf. $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 40 \times \dfrac{4.5 \times 10^{-4}\ \text{Wb}}{0.30\ \text{s}} = 0.060\ \text{V}$.
3. Current. $I = \dfrac{0.060\ \text{V}}{6.0\ \Omega} = 0.010\ \text{A}$.
4. Charge. $Q = \dfrac{N\Delta\Phi}{R} = \dfrac{40 \times 4.5 \times 10^{-4}\ \text{Wb}}{6.0\ \Omega} = 3.0 \times 10^{-3}\ \text{C}$.

Practise with new numbers

Check yourself

1. The flux linkage of a coil grows at a steady rate. The rate is then made three times as large. What happens to the emf?

2. A loop in the page is in a field that points into the page. The field weakens. Which way does the induced current turn?

3. A rod $0.50$ m long moves at $6.0$ m/s at right angles to its length and to a field of $0.12$ T. What is the emf between its ends?