11.2 Faraday’s law of induction: lenz’s law  (Page 3/6)

Calculating emf: how great is the induced emf?

Calculate the magnitude of the induced emf when the magnet in [link] (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 $B\text{cos}\theta$ (this is given, since the bar magnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.

Strategy

To find the magnitude of emf, we use Faraday’s law of induction as stated by $\text{emf}=-N\frac{\mathrm{\Delta }\Phi }{\mathrm{\Delta }t}$ , but without the minus sign that indicates direction:

Solution

We are given that $N=1$ and $\mathrm{\Delta }t=0\text{.}\text{100}\text{s}$ , but we must determine the change in flux $\mathrm{\Delta }\Phi$ before we can find emf. Since the area of the loop is fixed, we see that

Now $\mathrm{\Delta }(B\text{cos}\theta )=0\text{.}\text{200 T}$ , since it was given that $B\text{cos}\theta$ changes from 0.0500 to 0.250 T. The area of the loop is $A={\mathrm{\pi r}}^2=(3\text{.}\text{14}\text{.}\text{.}\text{.})(0\text{.}\text{060 m}{)}^2=1\text{.}\text{13}\times {\text{10}}^{-2}{\text{m}}^2$ . Thus,

Entering the determined values into the expression for emf gives

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.