Turning a coil in a field
A power station, a car's alternator and a bicycle dynamo all do one thing. Each turns a coil and a magnet past each other and collects the emf that the changing flux induces. A machine that does this with a steady turning is an AC generator. Its turning coil is called the armature.
Take a flat coil of \(N\) turns, each of area \(A\), in a uniform field \(\vec B\). Let \(\theta\) be the angle between the normal to the coil and the field, as for the magnetic moment of a loop. The flux through one turn is \(BA\cos\theta\). The flux linkage, the flux through all the turns together, is \(N\) times that.
Now let the coil turn at a steady angular speed \(\omega\). Start the clock when the normal lies along \(\vec B\), so that \(\theta = \omega t\). The flux linkage then rises and falls as a cosine: greatest, \(NBA\), at \(t = 0\), zero a quarter turn later, and \(-NBA\) after half a turn.
The alternating emf
The emf is minus the rate of change of the flux linkage. Differentiating a cosine gives a sine, so the emf alternates too. Its greatest value, reached when \(\sin\omega t = \pm1\), is the peak emf \(\varepsilon_0 = NBA\omega\).
\[ N\Phi = NBA\cos\theta \]
Why. the flux through each turn is \(BA\cos\theta\), and \(N\) turns carry it.
\[ \theta = \omega t \]
Why. the coil turns steadily, and the clock starts with the normal along \(\vec B\).
\[ \varepsilon = -\dfrac{d(N\Phi)}{dt} = -NBA\,\dfrac{d}{dt}\cos\omega t \]
Why. Faraday's law for a coil of \(N\) turns.
\[ \varepsilon = NBA\omega\sin\omega t = \varepsilon_0\sin\omega t \]
Read it. the derivative of \(\cos\omega t\) is \(-\omega\sin\omega t\), and the two minus signs cancel.
Watch Fig 2.1. When the flux linkage is at its peak, \(0.5\,\text{Wb}\), the emf is zero. When the flux crosses zero, the emf is at its peak. The emf is not set by how much flux the coil holds, but by how fast that flux changes. The current in each side reverses every half turn, so the output is an alternating emf. At \(2\omega\) the peak emf doubles from \(0.79\) to \(1.57\,\text{V}\), and the period halves from \(4\) to \(2\,\text{s}\).
The same emf follows from the moving wires. Let the coil be a rectangle with sides \(l\) along the axis and \(b\) across it. Each side of length \(l\) moves at \(v = \omega b/2\) on a circle. Only the part of its velocity across \(\vec B\), \(v\sin\omega t\), gives a motional emf \(Blv\sin\omega t\). The two sides are in series and their emfs add, giving \(2Bl(\omega b/2)\sin\omega t = BA\omega\sin\omega t\) for each turn.
A generator coil has \(200\) turns, each of area \(0.05\,\text{m}^2\), in a field of \(0.2\,\text{T}\), and turns at \(50\,\text{Hz}\). Find (a) the peak emf, (b) the rms emf and (c) the emf \(\tfrac{1}{600}\,\text{s}\) after the plane of the coil was across the field.
(a) \(\omega = 2\pi\times50 = 100\pi\,\text{rad/s}\), so \(\varepsilon_0 = NBA\omega = 200\times0.2\times0.05\times100\pi = 200\pi \approx 628\,\text{V}\).
(b) For a sine wave the rms value is the peak divided by \(\sqrt2\): \(628/\sqrt2 \approx 444\,\text{V}\). The next chapter, Alternating Current and the LCR Circuit, explains the rms value.
(c) With the plane across the field, the normal lies along it, so the clock starts there. Then \(\omega t = 100\pi/600 = \pi/6\), and \(\varepsilon = 628\sin 30° \approx 314\,\text{V}\).
Values used: \(\pi\), \(\sqrt2 = 1.414\) and the numbers given.
A coil of \(100\) turns, each of area \(0.02\,\text{m}^2\), turns at \(3000\) revolutions per minute in a field of \(0.5\,\text{T}\). What is its peak emf, in volts?
The generator of the last check has a peak emf of \(314\,\text{V}\). What is its emf when the coil has turned \(30°\) from the position of greatest flux, in volts?
Slip rings and brushes
The coil turns, but the circuit it feeds stays still. If its ends were fixed to wires, the wires would twist and break. So each end of the coil is joined to its own metal ring on the shaft, and the rings turn with the coil. These are the slip rings. Two carbon brushes press lightly on the rings and carry the current to the fixed circuit.
Fig 2.2 shows where the emf comes from. With the plane of the coil across the field, the two sides move along the field lines for an instant. They cut no lines, so there is no emf, although the flux is at its greatest. With the plane along the field, the sides cut straight across the lines. The emf is then greatest, although no flux passes through the coil at that instant.
At which position of a generator coil is its emf zero?
Real machines
A large power-station generator keeps its heavy coils still and turns an electromagnet inside them. Only the relative motion of coil and field matters, so the emf is the same. The fixed coils need no slip rings for the large current they carry. A bicycle dynamo works the same way: the tyre spins a small magnet inside a fixed coil. Ride faster, and \(\omega\) grows, so the lamp glows brighter.
A generator cannot give energy for nothing. When a current flows in its coil, the field pushes on the current in each side. By Lenz's law these forces oppose the turning. So the turbine must do work to keep the coil turning, and more current needs more work. That work is the electrical energy the generator delivers. A Delhi Metro train uses this in reverse when it brakes. Its motors run as generators, and the drag of the current they make slows the train while sending energy back to the line.
A bicycle dynamo lights a lamp. The lamp is switched off, so no current flows. What does the rider notice?
At the Tehri dam in Uttarakhand, water from the Bhagirathi falls through large pipes onto the turbines of the power house. Each turbine turns the electromagnet of a generator inside its fixed coils. The machines are held at a fixed speed, so their emf alternates exactly \(50\) times a second. Every generator on the Indian grid turns in step at this one frequency. A slow water turbine carries many pairs of poles, so each turn gives many cycles, and \(50\,\text{Hz}\) needs only a slow turning.
Four ways to double the peak
The peak emf \(NBA\omega\) is a product of four factors, so doubling any one of them doubles it. Wind twice the turns, use a field twice as strong, make each turn twice the area, or turn the coil twice as fast. Only the last one changes the frequency as well. The chip \(2\omega\) in Fig 2.1 showed it: the peak doubled and the period halved.
A coil of \(100\) turns, each of area \(0.05\,\text{m}^2\), turns in a field of \(0.1\,\text{T}\). Find its peak emf (a) at \(100\,\text{rad/s}\) and (b) at \(50\) revolutions per second.
(a) \(\varepsilon_0 = NBA\omega = 100\times0.1\times0.05\times100 = 50\,\text{V}\).
(b) \(\omega = 2\pi\times50 = 314\,\text{rad/s}\), so \(\varepsilon_0 = 100\times0.1\times0.05\times314 = 157\,\text{V}\). The emf now also alternates \(50\) times a second.
A generator gives a peak emf \(\varepsilon_0\). Which changes, made one at a time, give a peak of \(2\varepsilon_0\)?
Within a year of Faraday's discovery, the Paris instrument maker Hippolyte Pixii built a machine to use it. A hand crank turned a horseshoe magnet under two fixed coils of wire. As the poles swept past the coils, the current in them reversed every half turn: it alternated. The scientists of the day wanted a steady current for their experiments. So Pixii added a switch on the shaft that reversed the connections every half turn, the first commutator.
- Expecting the emf to be greatest when the flux is greatest.The emf follows the rate of change of flux. It is greatest when the flux is zero and zero when the flux is greatest.
- Measuring \(\theta\) from the plane of the coil.In \(NBA\cos\theta\), \(\theta\) is the angle between the normal and the field. With the plane across the field, \(\theta = 0\).
- Using the frequency in place of \(\omega\).The peak is \(NBA\omega\) with \(\omega = 2\pi f\). At 50 Hz, \(\omega = 314\,\text{rad/s}\).
- Thinking a generator makes energy.The current in the coil opposes the turning, so the turbine must supply all the energy delivered.