Magnetic flux and Faraday's law

A coil, a magnet and a meter

Join a coil of wire to a sensitive meter. There is no battery in the circuit. Now push a bar magnet into the coil. The needle of the meter jumps. Hold the magnet still inside the coil, and the needle goes back to zero. Pull the magnet out, and the needle jumps the other way.

Push the magnet in faster, and the jump is bigger. Keep the magnet still and move the coil instead, and you see the same jumps. So a magnetic field alone does nothing. A field that changes, as the coil sees it, makes a current. Michael Faraday found this in 1831. A current needs an emf to drive it, so we call this an induced emf.

What exactly is changing? We need a way to count the field that goes through a loop. That count is the magnetic flux.

Magnetic flux

Think of field lines as a stream of water. A loop held in the stream catches the most water when it faces the stream. It catches none when it is edge-on. The flux counts the field that goes through the loop in the same way.

For a flat surface of area $A$ in a uniform field $B$, the magnetic flux is

Here $\theta$ is the angle between the field and the normal to the surface. The normal is the line at right angles to the loop. Flux is a scalar. Its unit is the weber (Wb), and $1\ \text{Wb} = 1\ \text{T m}^2$.

The flux is greatest when the loop faces the field, with $\theta = 0$. It is zero when the field runs along the plane of the loop, with $\theta = 90^\circ$. It is negative for an angle between $90^\circ$ and $180^\circ$, because the field then enters the loop from the other side.

A coil of $N$ turns has the same flux $\Phi$ through each turn. The total, $N\Phi$, is the flux linkage.

A first example

A rectangular loop is $12$ cm long and $8.0$ cm wide. It sits in a field of $0.45$ T, and its plane makes an angle of $30^\circ$ with the field. Find the flux through the loop.

1. The area. $A = 0.12\ \text{m} \times 0.080\ \text{m} = 9.6 \times 10^{-3}\ \text{m}^2$.
2. The angle. The angle in the formula is the one between the field and the normal. The normal is at $90^\circ$ to the plane, so $\theta = 90^\circ - 30^\circ = 60^\circ$.
3. The flux. $\Phi = BA\cos\theta = 0.45\ \text{T} \times 9.6 \times 10^{-3}\ \text{m}^2 \times \cos 60^\circ = 2.16 \times 10^{-3}\ \text{Wb}$.

Turn the loop through $90^\circ$, and the flux falls to zero. Turn it on to $180^\circ$, and the flux is as large as at the start, but negative.

Three ways to change the flux

Look at $\Phi = BA\cos\theta$. The flux changes if any one of the three factors changes.

• The field $B$ changes. Move a magnet near the loop, or change the current in a coil close by.
• The area $A$ changes. Stretch the loop, squeeze it, or pull it out of the field.
• The angle $\theta$ changes. Turn the loop in the field. A generator works in this way.

A change of any one of them drives a current in a closed loop.

Faraday's law

Faraday's experiments say two things.

• First law. An emf is induced in a loop whenever the flux through it changes. It lasts only while the flux is changing.
• Second law. The size of the induced emf equals the rate of change of the flux linkage.

In symbols, for a coil of $N$ turns, the average emf over a time $\Delta t$ is

The exact form uses a very short time: $\varepsilon = -N\,d\Phi/dt$. The minus sign gives the direction of the emf, and Lenz's law explains it. For now we find only the size, $|\varepsilon| = N\,|\Delta\Phi|/\Delta t$.

Check the units. A weber is a tesla metre squared, and one weber each second is one volt. So the right side is in volts, as an emf must be.

Notice what the law does not say. A big flux does not give a big emf. A steady flux, however large, gives no emf at all. Only the rate of change of the flux counts.

A second example

A coil has $120$ turns, and each turn has an area of $25\ \text{cm}^2$. The coil faces a field that rises from $0.10$ T to $0.70$ T in $0.40$ s. The coil has a resistance of $9.0\ \Omega$. Find the emf, the current and the charge that flows.

1. The change of flux. For one turn, $\Delta\Phi = A\,\Delta B = 2.5 \times 10^{-3}\ \text{m}^2 \times 0.60\ \text{T} = 1.5 \times 10^{-3}\ \text{Wb}$.
2. The emf. $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 120 \times \dfrac{1.5 \times 10^{-3}\ \text{Wb}}{0.40\ \text{s}} = 0.45\ \text{V}$.
3. The current. We want $I$, so $I = \dfrac{\varepsilon}{R} = \dfrac{0.45\ \text{V}}{9.0\ \Omega} = 0.050\ \text{A}$.
4. The charge. $Q = I\,\Delta t = 0.050\ \text{A} \times 0.40\ \text{s} = 0.020\ \text{C}$.

Now let the same change happen four times faster, in $0.10$ s. The emf becomes $1.8$ V and the current becomes $0.20$ A. But the charge is $0.20\ \text{A} \times 0.10\ \text{s} = 0.020$ C, the same as before.

Choose a long time for the change, then a short one. The line gets steeper when the time is short. The emf is the steepness of this line, because the graph already shows the flux linkage $N\Phi$.

The charge does not depend on the time

Why did the charge stay the same? Put $I = \varepsilon/R$ and $\varepsilon = N\Delta\Phi/\Delta t$ into $Q = I\,\Delta t$:

The time cancels. The total charge depends only on the change of flux linkage and on the resistance. A fast change gives a large current for a short time. A slow change gives a small current for a long time. The charge is the same. We use this to measure magnetic fields, as one of the problems shows.

Practise with new numbers

Try it in Python

The program finds the flux through a coil at six angles. It then finds the average emf when the coil is turned from facing the field to $90^\circ$ and to $180^\circ$.

Common mistakes

• Using the angle with the plane. The angle in $\Phi = BA\cos\theta$ is the one between the field and the normal. A loop whose plane lies along the field has $\theta = 90^\circ$ and no flux.
• Forgetting the turns. The emf of a coil is $N$ times the emf of one turn.
• Mixing cm² and m². $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.
• Thinking that a big flux gives a big emf. The emf comes from the rate of change of the flux. A steady flux gives none.

Remember these

• Flux: $\Phi = BA\cos\theta$, in weber. $1\ \text{Wb} = 1\ \text{T m}^2$. Flux linkage: $N\Phi$.
• Faraday's law: $|\varepsilon| = N\dfrac{|\Delta\Phi|}{\Delta t}$. One weber each second is one volt.
• The flux changes when $B$, $A$ or $\theta$ changes.
• The current is $I = \varepsilon/R$, and the charge that flows is $Q = N\Delta\Phi/R$. The time of the change does not matter.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A loop of area $60\ \text{cm}^2$ faces a field of $0.20$ T head on, with the normal along the field. What is the flux through it?

2. A loop lies with its plane along a uniform field. What is the flux through it? Can an emf be induced if the loop is turned slowly?

3. The flux linkage of a coil falls from $0.80$ Wb to $0.20$ Wb in $0.30$ s. What is the average emf?

4. A strong magnet is held still inside a coil. A student says that there must be a large emf, because the flux is large. Is the student right?

5. A coil of resistance $20\ \Omega$ is closed, and the flux linkage through it changes by $0.060$ Wb. How much charge flows? How does the answer change if the change takes twice as long?