PHYS 208 Lecture 19

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Faraday's Law

${\mathcal {E}}=-{\frac {\mathrm {d} \Phi _{B}}{\mathrm {d} t}}$

Since Φ depends on $BA\cos {\phi }$, changing the magnetic field, the area, or the angle formed between the normal and magnetic field will produce an emf

Lenz's Law

The direction of the induced Emf is such that the induced current will produce a magnetic field which counters the 'change in flux'. So if the flux is decreasing (thus dΦ/dt is negative), then the induced emf will try to increase B to maintain Φ. If the flux is increasing (positive dΦ/dt), then induced emf will try the prevent Φ from increasing.

Motional Emf

Moving any material through a magnetic field produces an emf. If the material is a conductor, the charges will congregate at opposites sides, producing an induced electric field.

Example

Moving a bar of length $L$ with a constant velocity $v$ through a uniform magnetic field $B$:

${\begin{aligned}{\mathcal {E}}&=\int _{0}^{L}({\vec {v}}\times {\vec {B}})\cdot \mathrm {d} {\vec {\ell }}\\&=vB\int _{0}^{L}\mathrm {d} \ell \\&=vBL\end{aligned}}$

Application

An airplane is moving 200 m/s in the earth's magnetic field (approx. 4E-4 T). What is the emf induced between the wingtips (a distance of 10 m) due to motional emf?

${\mathcal {E}}=vBL=(200)(4\times 10^{4})(10)=.8\mathrm {V}$

E-fields in Space

Suppose we take an ideal solenoid and vary the current through it, there is a loop of wire around the solenoid (but not touching it) since the flux is changing, there will be and induced emf on the loop of wire.

Since emf is just a voltage, the integration definition still holds:

$V=\oint {\vec {E}}\cdot \mathrm {d} {\vec {\ell }}$

As the flux changes, we get ${\mathcal {E}}=\oint {\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-{\frac {\mathrm {d} \Phi _{B}}{\mathrm {d} t}}$, thus an induced electric field is not conservative!

Transformers

File:Transformer.png

Transformer

Since conductors (iron, for example) has a higher permeability, magnetic field lines do not like to leave the material.

Running a time-changing current through a wire wrapped around one side of a torus of iron will produce a time-changing magnetic field. This field produces an induced current on the wire wrapped around the other side. Varying the number of coils around the iron core changes the voltage of each side.

the "input" side of the transformer is called the primary side, has $N_{p}$ turns, and has a voltage of $V_{p}$. The opposite side is called the secondary side, has $N_{s}$ turns, and will produce a voltage of $V_{s}$:

${\frac {{\mathcal {E}}_{p}}{N_{p}}}={\frac {{\mathcal {E}}_{s}}{N_{s}}}$

Note: When measuring voltage of alternating current, V is the amplitude (from centerline to peak or trough). Another way is to measure root mean square (rms) voltage by squaring the voltage, taking the average value, and taking the square root of that.

Eddy Currents

A magnetic field is applied to a small section of a rotating conducting disk. As the charge carriers move through the magnetic field, they feel a force of $q{\vec {v}}\times {\vec {B}}$

This produces a current inside the magnetic field. Therefore, there is another fore inside the conductor $I\mathrm {d} {\vec {\ell }}\times {\vec {B}}$ that wants to slow the rotation.

I want this kind of braking system on my bike: no moving parts to wear down, and no squeaking!

Displacement Current

What happens to the magnetic field between the plates of a capacitor?

Generally, an amperian surface is a "drumhead" that is stretched tightly between the boudary. However, what if we make a surface that extends past one of the electrodes of the capacitor?

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=0$ between the plates.

BUT

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}I_{en}$ around the wire.

We get two different values for the same surface... What's going on here?

We have a time-changing electric charge build-up on the plates, which produces a time-changing electric field between the plates.

For parallel plates, $E={\frac {q}{A\epsilon _{0}}}$ by Gauss' law, so the changing electric field gives

${\frac {\mathrm {d} E}{\mathrm {d} t}}={\frac {1}{A\epsilon _{0}}}{\frac {\mathrm {d} q}{\mathrm {d} t}}={\frac {I}{A\epsilon _{0}}}$

Solving this for $I$ gives

$I=\epsilon _{0}{\frac {\mathrm {d} E}{\mathrm {d} t}}A=\epsilon _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}$

This current is called Displacement Current

Therefore, Ampere's law can be modified to be:

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}(I_{real}+I_{displacement})=\mu _{0}\left(I+\epsilon _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}\right)$

Maxwell's Equations

This is a very exciting time: the unification of electricity and magnetism into electromagnetism

$\oint {\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {q_{en}}{\epsilon _{0}}}$

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {A}}=0$

$\oint {\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-{\frac {\mathrm {d} \Phi _{m}}{\mathrm {d} t}}$

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}\left(I+\epsilon _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}\right)$

We will come back to these in our discussion on electromagnetic waves.

Chapter 30: Inductance

Now we introduce a new circuit element based on Faraday's law: the inductor

An inductor is nothing more than a coil of wire inside a circuit that store energy in the form of magnetic fields.

Two kinds of inductance:

mutual inductance

current flowing through one circuit affects the current in another circuit

self inductance

inductors resist changes in current due to their magnetic fields.