Question

An elastic conducting material is stretched into a circular loop of 10.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 60.0 cm/s. What emf is induced in the loop at that instant?

Answer 1

The area of a circle is $A = \pi r^2$, and that means that a small change in the area is given by:

$$\mathrm{dA} = 2 \pi \mathrm{r} \, \mathrm{dr}.$$ $$\mathrm{dA}/\mathrm{dt} = 2 \pi \mathrm{r} \, \mathrm{dr}/\mathrm{dt}.$$

This might be obvious after you think about it, since $2\pi$ is just the diameter of the circle.

So you've got a hoop with a magnetic field B going through it. That means there's a flux through that field:

$$\Phi = B A.$$

This flux is changing in time! Its derivative is:

$$\mathrm{d}\Phi/\mathrm{dt} = B \, \mathrm{dA}/\mathrm{dt} + A \, \mathrm{dB}/\mathrm{dt}$$

Simple calculus, yes? But if B isn't changing, then $\mathrm{dB}/\mathrm{dt} = 0$, and:

$$\mathrm{d}\Phi/\mathrm{dt} = B \, \mathrm{dA}/\mathrm{dt} = 2 \pi B \, \mathrm{r} \, \mathrm{dr}/\mathrm{dt}.$$

There has been very little physics, here, just calculus. But the physics is very easy! What does Faraday's law say about $\mathrm{d}\Phi/\mathrm{dt}$? That an EMF is produced in the ring equal exactly to $\mathrm{d}\Phi/\mathrm{dt}$.

$$\mathcal{E} = \frac{\mathrm{d}\Phi}{\mathrm{dt}} = 2 \pi B \, \mathrm{r} \, \frac{\mathrm{dr}}{\mathrm{dt}} = 2 \cdot 3.14 \cdot 0.8 \cdot 0.1 \cdot 0.6 = 0,3\,\mathrm{V}$$