PHY294H - Lecture 25

Recommended reading: FGT 852-3
Recommended problems: 31.45,31.46

Rotating loops - Electric Generators and motors

We will now show that if we rotate a loop at constant angular speed in a uniform and constant magnetic field, then we generate an oscillatory emf. In an electric generator, we rotate a loop through a magnetic field using a chemical or nuclear energy source. This produces the oscillatory emf we use in our society. An electric motor works in reverse. If we apply an oscillatory emf to a current loop in a magnetic field, it rotates at constant angular speed. Recall that in the case of a DC current source, we had to switch the current direction every $\pi$rotation in order to keep the loop moving in the same direction. However in the case of an AC motor, the AC source switches polatity automatically every $\pi$ rotation, so the direction of rotation of the loop is maintained.

Let us consider the situation in which a coil with N turns and area Ais rotated about its central axis at constant angular speed $\omega$. The angle between the magnetic moment of the loop and the applied magnetic field is $\psi$. This angle increases as $\psi(t) = \omega t$ due to the constant angular speed of the coil. The rate of change of the magnetic flux is

$${d \phi_B \over dt} = {d \over dt}(B A cos(\psi) ) = -B A \omega \ sin(\omega t)$$ (1)

The induced emf is given by Faraday's which gives,

$${\cal{E}} = - N {d\phi_B \over dt} = NBA \omega\ sin(\omega t)$$ (2)

The potential energy is given by,

$$U = -\vec{\mu}\cdot \vec{B} = NIAB\ cos(\psi)$$ (3)

The torque is given by,

$$\tau = \vec{\mu}\wedge\vec{B} = NIAB\ sin(\psi)$$ (4)

The mechanical power that must be supplied to rotate the coil is,

$$P_{mech} = \vec{\tau} \cdot \vec{\omega} = NIAB\omega \ sin(\psi)$$ (5)

This must be equal to the electrical power delivered to the load, this is given by,

$$P_{elec} = {{\cal{E}}^2 \over R} = {(NBA \omega)^2\over R} sin^2(\omega t)$$ (6)

The average power dissipated is

$$P_{av} = {1\over 2\pi} \int_0^{2\pi} {(NBA \omega)^2\over R} sin^2(\theta)d\theta$$ (7)

Using the result that,

$${1\over 2\pi} \int_0^{2\pi} sin^2(\theta)d\theta = {1\over 2}$$ (8)

we have,

$$P_{av} = {1\over 2} {(NBA \omega)^2\over R}$$ (9)

which is the average power delivered by our generator. This is the number you see quoted on your lightbulb.