If you need a pdf version of these notes you can get it here

The 2014 video was corrupted again. I suspect it is actually to do with running two cameras at once rather than the giant magnet…but I'm not totally sure…here is the 2013 video.

It is important to remember that an emf is not a force, but rather a measure of the work done in a circuit, so we can write the emf in terms of an integral over a closed path of the electric field

$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}$

and then

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$

Here we are taking an integral around the path that encloses the area in which magnetic flux is changing.

We should note that the implication of this is that in the presence of a time varying magnetic field the electric force is no longer a conservative force.

Suppose I want to find the field as function of radius inside a cylindrical electromagnet which experiences a time varying magnetic field. Looking down the bore of the magnet we can approximate the field as spatially uniform and directed along the axis of the electromagnet.

We first deduce the direction of the electric field using Lenz's law and then find the magnitude of the current using Faraday's Law.

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}\to E(2\pi r)=\pi r^{2}\frac{dB}{dt}$ for $r<R$

giving

$E=\frac{r}{2}\frac{dB}{dt}$

Once we are outside the solenoid however the flux starts to decrease as the area considered increases

$E(2\pi r)=\pi R^{2}\frac{dB}{dt}$ for $r>R$

and

$E=\frac{R^{2}}{2r}\frac{dB}{dt}$ for $r>R$

We saw in previous lectures that a torque could be produced on a current carrying loop placed in a magnetic field and that when appropriate commutation was used an electric motor could be made.

If we now think about it the other way round and consider the change in flux on a loop as it is rotated by some external torque we can see that it should produce some emf which can be used as the basis for an electric generator.

$\mathcal{E}=\frac{d\Phi_{B}}{dt}=-\frac{d}{dt}\int\vec{B}\cdot d\vec{A}=-\frac{d}{dt}BA\cos\theta$

If the loop rotates with a constant angular velocity $\omega=\frac{d\theta}{dt}$ then $\theta=\theta_{0}+\omega t$ and we can say that

$\mathcal{E}=-BA\frac{d}{dt}(\cos\omega t)=BA\omega\sin\omega t$

of course if there are $N$ loops

$\mathcal{E}=-NBA\frac{d}{dt}(\cos\omega t)=NBA\omega\sin\omega t=\mathcal{E}_{0}\sin\omega t$

If we want to produce a DC voltage we need to use a commutator. A simple split ring commutator will produce a time varying voltage, though it will have only one sign. if we add multiple armatures (additional coils at different angles), their sum will approach closer to a steady DC voltage. The signal can be smoothed by placing a capacitor in parallel (a low pass filter).

As an electric motor accelerates under the action of magnetic force the rate of change of flux $\frac{d\Phi_{B}}{dt}$ will get larger, producing an emf which opposes the motion. The total potential driving the current in the circuit is the sum of the applied potential and the back emf generated by the changing magnetic flux, which is in in the opposite direction to the applied potential. When these two are equal the current flowing through the circuit is zero. If there is no other load on the motor the motor would then turn at that speed constantly. Of course any load will reduce the speed, and hence the back emf, so at the constant top speed of a real motor there will still be some current flowing.

In a generator the opposite occurs, if the emf generated is used to produce a current then this produces a counter torque that opposes the motion.

We can compare the behavior of magnets above cooled copper and a high temperature superconductor and see that magnetic braking and levitation by superconductors are both products of magnetic induction.

In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same. The iron core is usually layered with insulating materials to prevent eddy currents. When an AC voltage is applied to the primary coil the magnetic flux passing through it is related to the applied field by

$V_{P}=N_{P}\frac{d\Phi_{B}}{dt}$

if we assume the coil has no resistance. The emf produced by the changing flux is $\mathcal{E}=-N_{P}\frac{d\Phi_{B}}{dt}$ and this means that the net potential drop in the coil is actually zero (as Kirchoff's rules say it must be if the resistance of the coil is zero). The voltage induced in the secondary coil will have magnitude

$V_{S}=N_{S}\frac{d\Phi_{B}}{dt}$

We can thus see that

$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$

If we assume there is no power loss (which is fairly accurate) then $I_{P}V_{P}=I_{S}V_{S}$ and

$\frac{I_{S}}{I_{P}}=\frac{N_{P}}{N_{S}}$