132 Electric Generators
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Learning Objectives
- Calculate the emf induced in a generator.
- Calculate the peak emf which can be induced in a particular generator system.
Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux. We will now explore generators in more detail. Consider the following example.
The generator coil shown in (Figure) is rotated through one-fourth of a revolution (from \(\theta =0º\) to \(\theta =\text{90º}\) ) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?
Strategy
We use Faraday’s law of induction to find the average emf induced over a time \(\Delta t\):
We know that \(N=\text{200}\) and \(\Delta t=\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{ms}\), and so we must determine the change in flux \(\Delta \Phi \) to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that
Now, \(\Delta \left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=-1\text{.}0\), since it was given that \(\theta \) goes from \(\text{0º}\) to \(\text{90º}\) . Thus \(\Delta \Phi =-\text{AB}\), and
The area of the loop is \(A={\mathrm{\pi r}}^{2}=\left(3\text{.}\text{14}\text{.}\text{.}\text{.}\right)\left(0\text{.}\text{0500}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}=7\text{.}\text{85}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\). Entering this value gives
Discussion
This is a practical average value, similar to the 120 V used in household power.
The emf calculated in (Figure) is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width \(w\) and height \(\ell \) in a uniform magnetic field, as illustrated in (Figure).
Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be \(\text{emf}=\mathrm{B\ell v}\), where the velocity v is perpendicular to the magnetic field \(B\). Here the velocity is at an angle \(\theta \) with \(B\), so that its component perpendicular to \(B\) is \(v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) (see (Figure)). Thus in this case the emf induced on each side is \(\text{emf}=\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \), and they are in the same direction. The total emf around the loop is then
This expression is valid, but it doe