Example 23.1 from College Physics, 23.2 Faraday’s Law of Induction: Lenz’s Law

Calculating Emf: How Great Is the Induced Emf?

Calculate the magnitude of the induced emf when the magnet in Figure 23.7(a) is thrust into the coil, given the following information: the single loop coil has a radius of 6.00 cm and the average value of $B cos θ$ (this is given, since the bar magnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.

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
To find the magnitude of emf, we use Faraday’s law of induction as stated by $emf = - N \frac{Δ Φ}{Δ t}$ , but without the minus sign that indicates direction:
$emf = N \frac{Δ Φ}{Δ t} .$
23.3
Solution
We are given that $N = 1$ and $Δ t = 0 . 100$ , but we must determine the change in flux $Δ Φ$ before we can find emf. Since the area of the loop is fixed, we see that
$Δ Φ = Δ (BA cos θ) = A Δ (B cos θ).$
23.4

Now $Δ (B cos θ) = 0 . 200 T$ , since it was given that $B cos θ$ changes from 0.0500 to 0.250 T. The area of the loop is $A = {πr}^{2} = (3.14 . . .) (0 . 060 m)^{2} = 1 . 13 \times 10^{- 2} m^{2}$ . Thus,
$Δ Φ = (1.13 \times 10^{- 2} m^{2}) (0.200 T).$
23.5

Entering the determined values into the expression for emf gives
$Emf = N \frac{Δ Φ}{Δ t} = \frac{(1.13 \times 10^{- 2} m^{2}) (0.200 T)}{0.100 s} = 22 . 6 mV.$
23.6
Discussion
While this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in the coil, a stronger magnet, and faster movement make induction the practical source of voltages that it is.

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Strategy

Solution

Discussion

Strategy

Solution

Discussion