Many old bicycles in India carry a small dynamo pressed against the tyre. It feeds a front lamp with no battery at all. Inside the dynamo a small magnet spins inside a coil of wire. When the cycle stands still, the lamp is dark, although the magnet sits right inside the coil. Start pedalling and the lamp glows, and it glows brighter as you go faster.
A changing flux drives a current
Join a coil to a sensitive centre-zero meter, a galvanometer, with no cell in the circuit. Hold a bar magnet still near the coil. The needle does not move. Now push the magnet towards the coil. The needle kicks one way while the magnet moves and returns to zero when it stops. Pull the magnet away, and the needle kicks the other way.
Watch the two graphs in Fig 2.1. The flux rises only while the magnet moves, and the emf is non-zero only then. The emf is largest where the flux graph is steepest, and zero where it is flat, even at its highest value. Choose "twice as fast": the flux rises to the same \(0.1\,\text{mWb}\) in half the time. The emf is twice as large, about \(43\,\text{mV}\) instead of \(21\,\text{mV}\). The emf follows the rate of change of the flux, not the flux itself.
An emf that appears because a flux changes is an induced emf, and the process is electromagnetic induction. The emf acts like a cell spread through the circuit. If the circuit is closed, it drives an induced current \(I = \varepsilon/R\), where \(R\) is the total resistance. The idea of emf and of \(I = \varepsilon/R\) is the one met in the chapter Electric Current and Resistance.
The emf induced in a circuit equals the rate of change of the magnetic flux linked with it. For a coil of \(N\) turns, \(\varepsilon = -N\,\dfrac{d\Phi}{dt}\). The minus sign says that the emf opposes the change of flux; that is Lenz's law, the subject of the next lesson.
Now the bicycle lamp is clear. A magnet at rest inside the dynamo's coil gives a large but steady flux. So the emf is zero and the lamp stays dark. Pedalling spins the magnet, and the flux through the coil swings back and forth. Pedal twice as fast and the same swings happen in half the time. The emf doubles, and the lamp glows brighter.
A coil of \(200\) turns, each of area \(0.01\,\text{m}^2\), lies with its axis along a field of \(0.2\,\text{T}\). The field falls to zero in \(0.1\,\text{s}\). Find (a) the average emf. A single loop of area \(0.5\,\text{m}^2\) and resistance \(2\,\Omega\) lies across a field rising at \(0.4\,\text{T/s}\). Find (b) the emf, the current and the electrical power. A flexible loop in a field of \(0.5\,\text{T}\) is squeezed from \(0.2\,\text{m}^2\) to \(0.1\,\text{m}^2\) in \(0.5\,\text{s}\). Find (c) its average emf.
(a) \(\Delta\Phi = 0.2 \times 0.01 = 2\times10^{-3}\,\text{Wb}\), so \(\varepsilon\) \(= N\Delta\Phi/\Delta t\) \(= 200 \times 2\times10^{-3}/0.1\) \(= 4\,\text{V}\).
(b) \(\varepsilon = A\,dB/dt = 0.5 \times 0.4 = 0.2\,\text{V}\). Then \(I = 0.2/2 = 0.1\,\text{A}\) and \(P = \varepsilon I = 0.02\,\text{W}\). Whatever raises the field supplies this power.
(c) \(\Delta\Phi = B\,\Delta A = 0.5 \times 0.1 = 0.05\,\text{Wb}\), so \(\varepsilon = 0.05/0.5 = 0.1\,\text{V}\).
Values used: only the data; the sizes of the emfs are asked, and their senses come in the next lesson.
A coil of \(500\) turns, each of area \(4\times10^{-3}\,\text{m}^2\), has its axis along a field of \(0.3\,\text{T}\). The field is switched off and falls to zero in \(0.05\,\text{s}\). What is the average emf, in volts?
A strong bar magnet rests inside a coil joined to a galvanometer. What does the galvanometer read?
The emf is the slope
On a graph of flux against time, \(d\Phi/dt\) is the slope of the tangent, as the velocity is the slope of a position-time graph. So the emf at any moment is \(-N\) times that slope. For a fixed coil of area \(A\) in a changing field, \(\Phi = BA\). The emf is then \(-NA\) times the slope of the graph of \(B\) against \(t\).
In Fig 2.2 the slope in stage 1 is \(0.4/2 = 0.2\,\text{T/s}\), so \(\varepsilon = -50 \times 0.1 \times 0.2 = -1\,\text{V}\). In stage 2 the graph is flat, and the emf is zero although the field is at its largest. In stage 3 the slope is \(-0.4\,\text{T/s}\), twice as steep and downhill. The emf is \(+2\,\text{V}\), twice as large and in the opposite sense.
The flux through a metal ring varies as \(\Phi = 6t^3 - 18t^2\), in webers, with \(t\) in seconds. The ring has a resistance of \(3\,\Omega\). Find the largest induced current between \(t = 0\) and \(t = 2\,\text{s}\).
The emf is \(-d\Phi/dt = -(18t^2 - 36t)\), and the current is its size divided by \(3\,\Omega\).
\(I\) \(= |18t^2 - 36t|/3\) \(= |6t^2 - 12t|\), in amperes.
\(d(12t - 6t^2)/dt = 12 - 12t = 0\) at \(t = 1\,\text{s}\). Then \(I = 12 - 6 = 6\,\text{A}\).
The largest current is \(6\,\text{A}\), at \(t = 1\,\text{s}\), when the emf is \(18\,\text{V}\).
The field through a coil of \(20\) turns, each of area \(0.05\,\text{m}^2\), rises steadily from zero to \(0.6\,\text{T}\) in \(3\,\text{s}\). What is the size of the emf during the rise, in volts?
The charge depends only on the change
A fast change gives a large current for a short time. A slow change gives a small current for a long time. How much charge flows in all? Add up \(I\,dt\) over the change.
\[ I = \dfrac{|\varepsilon|}{R} = \dfrac{N}{R}\left|\dfrac{d\Phi}{dt}\right| \]
Why. Faraday's law gives the emf, and the current is the emf over the total resistance \(R\).
\[ dq = I\,dt = \dfrac{N}{R}\,|d\Phi| \]
Why. in a short time \(dt\) the charge \(I\,dt\) passes; the \(dt\) cancels against the \(dt\) of the rate.
\[ q = \dfrac{N}{R}\int |d\Phi| = \dfrac{N\,|\Delta\Phi|}{R} \]
Read it. add over the whole change, with the flux changing one way throughout.
The time has dropped out. For the \(200\)-turn coil of the first example, joined into a circuit of \(8\,\Omega\), the charge is \(200 \times 2\times10^{-3}/8 = 0.05\,\text{C}\). This is the same whether the field dies in a thousandth of a second or in a minute. A search coil uses this: flipped or pulled out of a field, it sends a charge \(q\) through a meter, and \(q\) gives the flux.
A coil of \(100\) turns, each of area \(0.02\,\text{m}^2\), lies face-on in a field of \(0.5\,\text{T}\). The circuit has a total resistance of \(5\,\Omega\). The coil is pulled right out of the field. How much charge flows, in coulombs?
A magnet is pushed into a coil once slowly and once quickly, from the same start to the same end. The galvanometer kicks further the second time. Explain why the charge that flows is nevertheless the same.
The emf is \(N\,d\Phi/dt\), so a quick push gives a larger emf and a larger current, which makes the bigger kick. But the current lasts for a shorter time. The charge is the current times the time, which adds up to \(N\Delta\Phi/R\). Both pushes give the same change of flux through the same resistance, so the same charge flows.
Turning a coil
Start with the normal of a coil along the field, flux \(+BA\) through each turn. Flip the coil over through \(180°\). Now the same lines cross it from the back, so the flux is \(-BA\). The change is \(2BA\) per turn, not \(BA\). Flipping a coil is the quickest way to get a large change of flux from a weak field.
A coil of \(100\) turns, each of area \(0.05\,\text{m}^2\), has its normal along a field of \(0.1\,\text{T}\). It is flipped through \(180°\) in \(0.1\,\text{s}\). Find (a) the average emf. A second coil of \(500\) turns, each of area \(0.1\,\text{m}^2\), starts with its normal along the Earth's field. The worksheet takes that field as \(5\times10^{-5}\,\text{T}\). The coil is flipped over in \(0.2\,\text{s}\). Find (b) its average emf.
(a) \(\Delta\Phi\) \(= 2BA\) \(= 2 \times 0.1 \times 0.05\) \(= 0.01\,\text{Wb}\), so \(\varepsilon = 100 \times 0.01/0.1 = 10\,\text{V}\).
(b) \(\Delta\Phi = 2 \times 5\times10^{-5} \times 0.1 = 10^{-5}\,\text{Wb}\), so \(\varepsilon\) \(= 500 \times 10^{-5}/0.2\) \(= 2.5\times10^{-2}\,\text{V}\), \(25\,\text{mV}\).
In Delhi the whole field is \(4.84\times10^{-5}\,\text{T}\), which gives \(24\,\text{mV}\) instead.
Values used: Delhi's field \(4.84\times10^{-5}\,\text{T}\); the worksheet's \(5\times10^{-5}\,\text{T}\) in (b).
Now let the coil turn steadily at an angular speed \(\omega\) about an axis in its own plane, at right angles to the field. Its normal starts along \(\vec B\), so the angle at time \(t\) is \(\omega t\). The flux linkage is \(N\Phi = NBA\cos\omega t\). Differentiating, the emf is \(\varepsilon = NBA\omega\sin\omega t\), a sine wave of peak \(NBA\omega\).
The emf is largest when the flux passes through zero, with the coil edge-on to the field. It is zero when the flux is largest, with the coil face-on. A coil of \(100\) turns of \(0.05\,\text{m}^2\) turning at \(100\,\text{rad/s}\) in \(0.1\,\text{T}\) has a peak emf of \(100 \times 0.1 \times 0.05 \times 100 = 50\,\text{V}\). This turning coil is the heart of the generator, built in the next chapter.
A coil that moves without turning through a uniform field is different. Carry it along at a steady velocity, its plane always at the same angle to the field. Its flux does not change, so no emf appears in it at all. Motion alone is not enough; the flux must change.
A coil of \(100\) turns, each of area \(0.05\,\text{m}^2\), turns at \(50\) revolutions per second in a field of \(0.1\,\text{T}\). Its axis of turning is at right angles to the field. What is its peak emf, in volts?
On 29 August 1831 Michael Faraday wound two coils of insulated wire on opposite sides of an iron ring. One coil went to a battery through a contact he could make and break. The other went to a wire running over a compass needle. When he made the contact, the needle swung and settled back. When he broke it, the needle swung the other way. A steady current in the first coil did nothing to the second. Within weeks he found the same kick by thrusting a magnet into a coil. A change was what counted.
A Delhi Metro smart card has no battery. Inside its plastic lies a flat coil of a few turns and a tiny chip. The reader at the gate drives a current that changes millions of times a second, and its field passes through the card held near it. The changing flux through the card's coil induces an emf that powers the chip for the moment the card is near. The chip then answers the reader, and the gate opens.
- Using the flux instead of its change.The emf is \(-N\,d\Phi/dt\). A large steady flux induces nothing; a small fast change can induce a lot.
- Taking the change as \(BA\) when a coil is flipped.Flipping through \(180°\) takes the flux from \(+BA\) to \(-BA\), a change of \(2BA\) per turn.
- Forgetting the number of turns.Each turn adds its own emf, so the coil's emf is \(N\) times that of one turn.
- Expecting more charge from a faster change.\(q = N\Delta\Phi/R\) does not contain the time. A faster change gives a larger current for a shorter time.