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Lecture 22 - Applications of Induction

Faraday's Law of Induction

Faraday's law of induction states that the induced emf in a circuit is equal to rate of change of magnetic flux through the circuit.

$\mathcal{E}=-\frac{d\Phi_{B}}{dt}$

If the circuit is made of a number of loops, $N$ that are wrapped so that the same flux passes through them all the net emf in the circuit is the sum of the emf from each loop

$\mathcal{E}=-N\frac{d\Phi_{B}}{dt}$

The negative sign in Faraday's law represents Lenz's law which states that

“An induced emf is always in such a direction as to oppose the change in flux causing it”

Lenz's Law

“An induced emf is always in such a direction as to oppose the change in flux causing it”

We can understand Lenz's law in terms of the current that flows through the circuit as result of the emf and the magnetic field it would produce.

The magnetic field produced by the emf current is an attempt to maintain the magnetic flux.

Lenz's law is a consequence of Newton's Third Law, the forces on the electrons in the conductor that produce the induced emf are the reaction forces to the whatever is causing the magnetic flux to change.

Qualitative aspects of Faraday's law and Lenz's law are summarized in the scenarios below.

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General form of Faraday's Law

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.

Electric field produced by time varying magnetic field

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

Then we 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 $r>R$

emf produced by an AC generator

We saw in previous lectures that a torque could be produced on a current carrying loop placed in a magnetic field.

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 $\omega=\frac{d\theta}{dt}$:

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

of course if there are $N$ loops

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

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DC generators

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).

Back emf and Counter Torque

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.

Magnetic braking and levitation

Recall the demo I showed you just before Spring Break. Here is a video version of it.

The idea of magnetic breaking is shown here.

Eddy currents need a circuit! See here.

With sufficiently strong eddy currents one can also levitate without superconductors: see here.

Transformers and Power Transmission

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 $\mathcal{E}=-N_{P}\frac{d\Phi_{B}}{dt}$.

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

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

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phy142kk/lectures/22.txt · Last modified: 2015/03/25 09:15 by kkumar
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