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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 [link] is rotated through one-fourth of a revolution (from $\theta =\mathrm{0\xba}$ to $\theta =\text{90\xba}$ ) 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 $\mathrm{\Delta}t$ :
We know that $N=\text{200}$ and $\mathrm{\Delta}t=\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{ms}$ , and so we must determine the change in flux $\mathrm{\Delta}\Phi $ to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that
Now, $\mathrm{\Delta}(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta )=-1\text{.}0$ , since it was given that $\theta $ goes from $\text{0\xba}$ to $\text{90\xba}$ . Thus $\mathrm{\Delta}\Phi =-\text{AB}$ , and
The area of the loop is $A={\mathrm{\pi r}}^{2}=(3\text{.}\text{14}\text{.}\text{.}\text{.})(0\text{.}\text{0500}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}=7\text{.}\text{85}\times {\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ . Entering this value gives
Discussion
This is a practical average value, similar to the 120 V used in household power.
The emf calculated in [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
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
Now, linear velocity $v$ is related to angular velocity $\omega $ by $v=\mathrm{r\omega}$ . Here $r=w/2$ , so that $v=(w/2)\omega $ , and
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