### Example: A Simple Alternator

A rectangular loop is made to rotate with constant angular speed $\omega$ about the axis shown. The magnetic field $\vec{B}$ is uniform and constant. Find the induced e.m.f.

The induced e.m.f. is given by:

$$\begin{aligned} \epsilon &=-\frac{d \Phi_{B}}{dt} \\ &=-\frac{d}{dt} \left( BA \cos{\omega t} \right) \\ &= \omega BA \sin{\omega t} \end{aligned}$$

### Example: Slidewire Generator

Consider a U-shaped conductor in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the figure, directed into the page.

A metal rod with length L across the two arms of the conductor is laid, forming a circuit, and moved to the right with constant velocity $\vec{v}$.

Find the induced e.m.f.

$$\begin{aligned} \epsilon &=-\frac{d \Phi_{B}}{dt} \\ &=-B \frac{dA}{dt} \\ &=-BLv \end{aligned}$$

**What if we do not want to use Faraday’s Law to compute the induced e.m.f.?** In order to do that, we shall let the resistance of the circuit at a given point in the slidewire’s motion be R. Notice that there is a force opposing the motion of the metal rod, due to the current and magnetic field. (You can find the direction of the opposing force by using the Fleming’s left hand rule or just $\vec{I} \times \vec{B}$)

The opposing force is given by:

$$\begin{aligned} F &= ILB \\ &= \frac{\epsilon}{R} LB \end{aligned}$$

The rate at which work must be done to move the rod through the magnetic field is given by:

$$\begin{aligned} P_{applied} &= Fv \\ &= \frac{\epsilon}{R} LBv \end{aligned}$$

Notice that the rate at which energy is dissipated in the circuit will be given by:

$$P_{dissipated} = \frac{\epsilon^{2}}{R}$$

Hence,

$$\begin{aligned} P_{applied} &= P_{dissipated} \\ \frac{\epsilon}{R}{LBv} &= \frac{\epsilon^{2}}{R} \\ \epsilon &= LBv \end{aligned}$$

We’re done.

Note that:

$$\epsilon = \left( \vec{v} \times \vec{B} \right) . \vec{L}$$

We will look at the above relation in the next post.

Next: Motional Electromotive Force