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Magnetic flux and Faraday’s law
Magnetic flux Φ = BA cos θ and the ways it can change, Faraday's experiments, Faraday's law of induction ε = −NΔΦ/Δt, reading emf from a flux–time graph, the emf in a rod sliding on rails, and mutual induction between two coils.
On 29 August 1831 Michael Faraday wound two separate coils of wire onto an iron ring and connected one of them to a galvanometer. When he connected the other coil to a battery, the galvanometer needle flicked and settled back to zero. When he disconnected it, the needle flicked the other way. A steady current did nothing. A changing one made electricity appear in a circuit that wasn't connected to anything. Every power station on Earth runs on what he saw that day.
Magnetic flux
Magnetic flux, Φ, measures how much magnetic field passes through an area. Picture it as the number of field lines threading a loop. For a flat area A in a uniform field B,
$$\Phi = B_{\parallel}A = BA\cos\theta$$
where θ is the angle between the field and the normal to the area (the line sticking straight out of the loop's face). $B_{\parallel} = B\cos\theta$ is the component of the field parallel to the normal, the part that actually passes through. The unit of flux is the weber (Wb); 1 Wb = 1 T m2. This is why B is sometimes called the magnetic flux density.
The flux through a loop changes if any of B, A or θ changes:
- B changes: a magnet moves closer to or further from the loop, or the current in a nearby electromagnet changes.
- A changes: the loop is squashed or stretched, or a rod slides along rails so the area of the circuit inside the field grows.
- θ changes: the loop rotates in the field, as in every generator.
- The loop moves into or out of a field: the area of the loop inside the field changes.
- B = 0.50 T, 10 cm loop, θ = 60°. Check Φ = BA cos θ = 2.5 × 10−3 Wb.
- With 100 turns, press Rotate to θ = 0° in 0.10 s. Check the average emf: ε = NΔΦ/Δt = 2.5 V.
- Now rotate to 90°. Why is the emf bigger than for the first rotation even though both took 0.10 s?
- Find two different angles that give exactly half the maximum flux. (Hint: cos θ = 0.5, and the loop could face either way.)
Faraday's observations
Push a bar magnet into a coil connected to a sensitive ammeter (galvanometer) and you see what Faraday saw:
- Magnet and coil both still: no current, however strong the magnet.
- Magnet moving into the coil: the needle deflects, but only while it moves.
- Magnet pulled out: the needle deflects the other way.
- Coil moved over a stationary magnet: the same result. Only the relative motion matters.
- Faster motion, a stronger magnet or more turns: a bigger deflection.
Faraday's law of induction
All of these are summed up in one law: an emf is induced in a circuit whenever the magnetic flux through it changes, and the size of the emf equals the rate of change of flux linkage.
$$\varepsilon = -N\frac{\Delta\Phi}{\Delta t}$$
ε is the induced emf in volts, N the number of turns and ΔΦ/Δt the rate of change of flux through each turn. NΦ is called the flux linkage. The minus sign is Lenz's law, which gives the direction (6F). Note that Faraday's law gives an emf; whether a current flows, and how much, depends on whether the circuit is complete and on its resistance, $I = \varepsilon/R$.
A coil of 200 turns, each of area 4.0 × 10−3 m2, sits face-on to a magnetic field that rises steadily from 0.10 T to 0.50 T in 0.20 s. Find the average induced emf, and the current if the coil's circuit has a resistance of 8.0 Ω.
$\Delta\Phi = \Delta B \times A = (0.50 - 0.10)(4.0 \times 10^{-3}) = 1.6 \times 10^{-3}$ Wb.
$|\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = 200 \times \dfrac{1.6 \times 10^{-3}}{0.20} = 1.6$ V.
$I = \dfrac{\varepsilon}{R} = \dfrac{1.6}{8.0} = 0.20$ A, flowing only during the 0.20 s.
A 50-turn square coil of side 8.0 cm is turned from face-on to edge-on in a 0.30 T field in 25 ms. Find the average emf.
Face-on: $\Phi_1 = BA\cos 0^\circ = (0.30)(0.080)^2 = 1.92 \times 10^{-3}$ Wb. Edge-on: $\Phi_2 = BA\cos 90^\circ = 0$.
$|\varepsilon| = 50 \times \dfrac{1.92 \times 10^{-3}}{0.025} = 3.8$ V.
Reading emf from a graph
Because ε = −NΔΦ/Δt, the emf at any moment is −N times the gradient of the flux–time graph. Where the flux is steady (flat), the emf is zero; where it rises steeply, the emf is large and negative; where it falls, the emf is positive.
- Press Push in at 0.50 m/s with 100 turns and note the peak emf. Now Pull out. What is different about the graph?
- Double the speed. By what factor does the peak emf change? What about the area under the emf–time graph? (The area is NΔΦ, the same for any speed.)
- Double the turns. Explain the change using flux linkage.
- Press Push right through. Why does the emf change sign as the magnet passes the middle of the coil?
- Watch the yellow label. When the magnet approaches, what pole does the coil make at its near end? Why does this make the magnet harder to push?
A conductor moving through a field
A straight rod of length l slides along two rails at speed v through a field B perpendicular to the rails. In time Δt the rod moves vΔt, so the area of the circuit inside the field increases by $lv\Delta t$ and the flux by $Blv\Delta t$. Faraday's law gives
$$\varepsilon = \frac{\Delta\Phi}{\Delta t} = Blv$$
You can also see it from the charges: the free electrons in the rod move with it at speed v, so each feels a magnetic force qvB along the rod. They pile up at one end, leaving the other end positive, until the electric field they create balances the magnetic force. The rod has become a source of emf.
A 0.40 m rod is pulled at a steady 3.0 m s−1 along rails through a 0.25 T field. The circuit resistance is 2.0 Ω. Find the emf, the current, the force needed to keep the rod moving and the power supplied.
$\varepsilon = Blv = (0.25)(0.40)(3.0) = 0.30$ V; $I = 0.30/2.0 = 0.15$ A.
The current in the rod feels a motor-effect force $F = lIB = (0.40)(0.15)(0.25) = 0.015$ N, opposing the motion (Lenz's law). To keep a steady speed you must pull with 0.015 N.
Mechanical power in: $Fv = 0.015 \times 3.0 = 0.045$ W. Electrical power out: $\varepsilon I = 0.30 \times 0.15 = 0.045$ W. The work you do becomes electrical energy, then heat in the resistance. Energy is conserved.
Mutual induction: a current in a second coil
Faraday's ring shows that you don't need a moving magnet. A changing current in one coil (the primary) makes a changing magnetic field, so the flux through a second coil nearby (the secondary) changes, and an emf is induced in it. This is mutual induction. A steady DC current in the primary produces a steady flux and no emf; switching it on or off produces a brief pulse; an alternating current produces a continuously changing flux and an alternating emf. An iron core between the coils concentrates the field so that more of the primary's flux passes through the secondary. This is exactly how a transformer works (6G), and how wireless phone chargers and induction cooktops transfer energy without wires.
Test yourself
A loop of area 0.020 m2 is in a 0.40 T field, with the normal to the loop at 60° to the field. The flux through the loop is
Φ = 0.40 × 0.020 × cos 60° = 4.0 × 10−3 Wb.
A bar magnet rests inside a coil connected to a galvanometer. The reading is
The flux is not changing, so no emf is induced.
The flux through a 100-turn coil falls uniformly from 6.0 × 10−3 Wb to 2.0 × 10−3 Wb in 0.050 s. The size of the induced emf is
ε = 100 × 4.0 × 10−3 ÷ 0.050 = 8.0 V.
On a graph of flux against time for a coil, the induced emf at a given moment is proportional to
ε = −NΔΦ/Δt, the rate of change of flux.
A 1.5 m metal axle on a truck moves at 25 m s−1 at right angles to a vertical field component of 5.0 × 10−5 T. The emf between its ends is
ε = Blv = 5.0 × 10−5 × 1.5 × 25 = 1.9 × 10−3 V.
Label the sliding rod diagram.
Will an emf be induced in the coil?
Magnetic flux is measured in . The flux through a loop is greatest when the field is parallel to the loop's . An emf is induced only when the flux is . The induced emf is proportional to the number of and to the rate of change of flux. A changing current in one coil inducing an emf in another is called induction.
A student pushes the north pole of a magnet into a coil connected to a galvanometer, holds it still inside, then pulls it out quickly. Describe and explain the galvanometer readings during each stage, with reference to Faraday's law and energy.
While the magnet is pushed in, the flux through the coil increases, so by Faraday's law (ε = −NΔΦ/Δt) an emf is induced and the galvanometer deflects. While the magnet is held still, the flux is large but constant, so ΔΦ/Δt = 0, no emf is induced and the reading returns to zero. When the magnet is pulled out, the flux decreases, so the emf has the opposite sign and the needle deflects the other way. Because it is pulled out quickly, the rate of change of flux is greater, so this deflection is larger (though briefer). The induced current produces a magnetic field that opposes the motion of the magnet in each case, so the student must do work to move it. This mechanical work is transformed into electrical energy in the coil, which becomes heat in the resistance of the circuit.