Example 23.3 from College Physics, 23.5 Electric Generators

Calculating the Emf Induced in a Generator Coil

The generator coil shown in Figure 23.20 is rotated through one-fourth of a revolution (from $θ = 0º$ to $θ = 90º$ ) 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?

The figure shows a schematic diagram of an electric generator. It consists of a rotating rectangular coil placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this coil are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The coil is attached to an axle with a handle at the other end. Outer ends of the two brushes are connected to the galvanometer. The axle is mechanically rotated from outside by an angle of ninety degree that is a one fourth revolution, to rotate the coil inside the magnetic field. A current is shown to flow in the coil in clockwise direction and the galvanometer shows a deflection to left.

Figure 23.20 When this generator coil is rotated through one-fourth of a revolution, the magnetic flux

Φ

changes from its maximum to zero, inducing an emf.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We use Faraday’s law of induction to find the average emf induced over a time $Δ t$ :

$emf = - N \frac{Δ Φ}{Δ t} .$
23.11

We know that $N = 200$ and $Δ t = 15 . 0$ , and so we must determine the change in flux $Δ Φ$ to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that

$Δ Φ = Δ (BA cos θ) = AB Δ (cos θ) .$
23.12

Now, $Δ (cos θ) = - 1 . 0$ , since it was given that $θ$ goes from $0º$ to $90º$ . Thus $Δ Φ = - AB$ , and
$emf = N \frac{AB}{Δ t} .$
23.13

The area of the loop is $A = {πr}^{2} = (3.14 . . .) (0.0500$ . Entering this value gives

$emf = 200 \frac{(7 . 85 \times 10^{- 3} m^{2}) (1.25 T)}{15 . 0 \times 10^{- 3} s} = 131 V.$
23.14
Discussion
This is a practical average value, similar to the 120 V used in household power.

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