23.5 Electric generators

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• 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.

Calculating the emf induced in a generator coil

The generator coil shown in [link] 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 $\Delta t$ :

$\text{emf}=-N\frac{\Delta \Phi }{\Delta t}\text{.}$

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

$\Delta \Phi =\Delta \left(\text{BA}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=\text{AB}\Delta \left(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)\text{.}$

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

$\text{emf}=N\frac{\text{AB}}{\Delta t}.$

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

$\text{emf}=\text{200}\frac{\left(7\text{.}\text{85}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(1\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}\text{T}\right)}{\text{15}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{131}\phantom{\rule{0.25em}{0ex}}\text{V.}$

Discussion

This is a practical average value, similar to the 120 V used in household power.

The emf calculated in [link] 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 [link] .

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 [link] ). 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

$\text{emf}=2\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{.}$

This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity $\omega$ . The angle $\theta$ is related to angular velocity by $\theta =\mathrm{\omega t}$ , so that

$\text{emf}=\text{2}\mathrm{B\ell v}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\omega t}\text{.}$

Now, linear velocity $v$ is related to angular velocity $\omega$ by $v=\mathrm{r\omega }$ . Here $r=w/2$ , so that $v=\left(w/2\right)\omega$ , and

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