Reading: Chapter 30, INDUCTANCE, Chapter 31, ALTERNATING CURRENT
Study guide: Chapter 31
So, the flux $\Phi$ of magnetic field through a single wire loop is proportional to the current flowing in the wire:$$i \propto \Phi$$.
Or we could just as easily write the relationship in terms of the total flux through $N$ loops of wire: $$i\propto N\Phi$$
We'll call the constant of proportionality $L$, the "inductance", and place it here...$$iL=N\Phi$$.
Taking derivatives of both sides...$$L\frac{di}{dt}=N\frac{d\Phi}{dt}$$
But wait a sec, whenever there's a change in flux, we get an emf...$${\cal E}=-N\frac{d\Phi}{dt}=L\frac{di}{dt}$$
[Example, $\ell=$10 cm long solenoid, circular cross section of radius 2 cm, wound with $n=$100 loops/cm=$10^4$ loops/m, using formula for $B_\text{middle}=\mu_0nI$.]
So, the flux through any one loop is $\Phi=BA = \mu_0nI*\pi r^2$.
The self-induction constant is$$\begineq L&=&N\Phi/I = \mu_0nNI\pi r^2/I\\ &=&\mu_0nNI\pi r^2/I = 4\pi^2*10^{-7}*10^4*1000*(0.02)^2\\ &=&4.0\times 10^{-4} = 0.4 mH \endeq$$in "milli-henries".
Use the loop rule for voltage: $$\sum_i \Delta V = 0$$
$$0=+V_B-V_R-{\cal E}=V_B-iR-L\frac{di}{dt}$$
Solution: $$i(t)=i_f\left(1-e^{t/\tau}\right).$$Where $i_f=V_B/R$ and $\tau=L/R$
blah
Warm up... using Lenz's law...
When current is flowing in $A$, what direction is $\myv B$ inside of circuit $B$?
Arguing as before with "self-inductance", we say in the situation below...
There is a magnetic field at 2 due to the current $i_1$ in loop 1.
The magnetic field $\myv B$ at 2 is proportional to $i_1$. The total flux through loop 2 is proportional to $\myv B$, and to the number of loops $N_2$, so again we have a proportionality relationship, and we'll put in a constant $M$, the "mutual inductance" $$i_1M=N_2\Phi_2$$
Taking derivatives, as in the case of self-inductance we find...$${\cal E}_2 = -M\frac{di_1}{dt}$$
What about a current in 2, $i_2$ influencing loop 1? The surprising result is that the constant for the influence going in the other direction is the same, so$$M=\frac{N_2\Phi_2}{i_1}=\frac{N_1\Phi_1}{i_2}.$$
A
loop of current has a magnetic dipole moment, $\myv \mu$ pointing
in a direction normal to the loop, in the $\uv n$ direction, of magnitude $\mu=I*A$.
On the atomic scale, electrons in 'orbit' around nuclei have a characteristic magnetic moment, depending on what orbital they're in,
We say a hunk of material is magnetized when it has a non-zero, total magnetic moment $\myv \mu_\text{total}$.
The magnetization, $\myv M$ is the total magnetic dipole per unit volume: $$\myv M = \frac{\myv \mu_\text{total}}{V}$$
If many magnetic dipoles are oriented all in the
same way, it looks like there should be a magnetic field in the direction of
$\myv \mu_\text{total}$. Now, make sense of this behaviour of magnets...
But if the dipoles
are all pointing in random directions??
When an external magnetic field is imposed on a chunk of matter, at least two responses are possible:
$\frac
{d M}{d B} \gt 0$: para-magnetism,
$\frac{d
M}{d B} \lt 0$: dia-magnetism.What follows is a model for one of these kinds of magnetism...figure out which one....
Imagine that a magnetic field along the $\uv z$ direction is turned on near a loop tilted by an angle $\theta$ below the $y$-axis as shown...
$\Rightarrow$ this is a model for a paramagnet: The current is fixed (quantum mechanics) but if the loop is free to re-orient, it will swing around such that $\myv \mu \parallel \myv B$.
In contrast, a diamagnet may be thought of as a material in which currents are induced (according to Lenz's Law) that oppose the field, so that the magnetic dipole moment is always opposite to $\myv B$.
In general, the total magnetic field $\myv B$ is the sum of the originally applied magnetic field $\myv B_0$ plus a contribution due to the magnetization of the material:$$\myv B = \myv B_0 + \mu_0 \myv M$$
For many materials, the magnetization is proportional to the final resulting field. The proportionality constant constant is called the magnetic susceptibility $\chi$,$$\myv M=\chi \myv B$$
$$U=-\myv \mu \cdot \myv B=\mu B \cos\theta$$
Some consequences
Where is the field above this electromagnet stronger? weaker?
What if you flip the magnet over?
Many common materials are weakly diamagnetic--including frogs:
Try this at home? Umm, well, you'll need something like the 16 T magnet at the University
of Nijmegen's High Field Magnet Laboratory.
A third major class of magnets are ferromagnets. These may have a magnetization even when the applied field is 0. These are what we mean when we talk of "permanent magnets" -- though we'll see that they're not really 'permanent'.
Why do magnets attract (unmagnetized) steel and iron?
Or if the other pole is brought close:
Now, if you take away the magnet the material is still
*kinda* magnetized. There is some remnant magnetization. Notice that
most of the little magnets have their dark ends pointing right.
How to get back to the originally, randomly ordered state??
De-magnetizing, or
Heating jostles the 'little magnets' around.
Above the Curie temperature a ferromagnet $\to$ paramagnetic. Without
an applied field, the magnetization $\to$ 0.
Nickel:
Curie temperature:
358 C
Melting point 1453 C
What do you think happens if you cool a material below the Curie temperature with an applied magnetic field?
An electro-magnet moves about a surface coated with some magnetic material, and gives it a remnant magnet field pointing one way or the other (these are the '1's and '0's of binary coding).
When a wire loop is rotated in a magnetic field, an emf is induced in the loop, according to an equation like...$$v(t)=V\cos \omega t$$
[What is the angular frequency for electric service in North America which alternates back and forth 60 times per second?]
We might describe the current in a similar fashion...$$i(t)=I\cos\omega t.$$
It is not terribly useful to talk about the average of a sinusoidally-varying function--why?
More useful that are sometimes positive and sometimes
negative is the "root-mean-square (rms) value": $$\begineq
V_\text{rms}
&=& \sqrt{\left
[I showed in class why $\left< \cos^2x\right>=\left< \sin^2x\right>=1/2$. ]
The simplest AC circuit, with only a resistor and an AC voltage source...
At any point in time:$$\begineq v(t)&=&i(t)R\\ V\cos \omega t &=& IR\cos \omega t.\endeq$$The voltage and current are in phase.
The power is $$p(t)=v(t) i(t)=VI\cos^2 \omega t$$...varies with time, not constant.
The average power is $$\begineq \left< p(t) \right> &=& VI\left<\cos^2\omega t\right> = VI(1/2)\\ &=&(V/\sqrt 2)(I/\sqrt 2)=V_\text{rms}I_\text{rms}=V^2_\text{rms}/R=I^2_\text{rms}R.\endeq$$
Some current flows, $I_1$ to create same emf in primary coil as $V_1$: $$v_1(t)={\cal E}_1(t) = N_1\frac{d\Phi}{dt}$$Here, $\Phi$ is the total flux through the primary--not the flux per loop.
A "soft iron" ferromagnetic core will acts like a flux 'conductor': Nearly all the flux passing through the primary coils stays inside the core, and arrives at the secondary coils.
On the secondary side, $$N_2\frac{d\Phi}{dt}=v_2.$$
The flux is the same, so solving for the change of flux and equating...$$\frac{v_1}{N_1}=\frac{v_2}{N_2}.$$
There are no resistances. We shall see (next week) that there is no power dissipation in an ideal inductor, and so any power dissipated in the right side (used, e.g. to cross a resistor across $v_2$) is the same as the power drawn on the left, so$$v_1i_1=v_2i_2$$
If a transmission line has some resistance in the transmission wires, how can we minimize the power loss during transmission?
$$p=i^2R$$
Voltage is stepped up from the generating station, eventually to 100s of thousands of volts.
It's stepped down to 7,200 V (neighborhood distribution) and then 240 V (household use). [How power grids work]
Remember the connection of a unit circle to trig functions...
Now, as long as the denominator for a particular function is not 0...
The kinds of time dependent functions that we've been discussing for voltages etc, like $A \cos (\omega t + \theta_0)$ can be thought of as...
Let's start by assuming that the current in our circuit with nothing but inductance is a sinusoidal:$$i=I\cos \omega t.$$
There will be an emf (voltage) across the inductor, resisting the change in flux$$\begineq{\cal E}(t)&=&-\frac{d\Phi}{dt}=-L\frac{di}{dt}\\ &=&-L\frac{d}{dt}I\cos \omega t=-LI\omega\sin(\omega t)\\ &=&LI\omega\cos(\omega t + \pi/2).\endeq$$
So the voltage phasor $v_L(t)\equiv{\cal E}(t)$ is $90^o$ "ahead" of the current phasor.
Power is still $v_L(t)i(t)$. Graphing
the product of two cos functions which are $\pi/2$ out of phase...
What
do you see as the average power??
For our circuit with a capacitor, we start again from the same current $$i=I\cos \omega t=\frac{dq}{dt}.$$
Integrate to get the charge on the capacitor: $$q(t)= \frac{I}{\omega}\sin\omega t$$
Since the voltage on a capacitor obeys $q=Cv_C$: $$v_C(t)= \frac{I}{C\omega}\sin\omega t = \frac{I}{C\omega}\cos(\omega t-\pi/2).$$
The phasor for capacitor voltage lags $90^o$ behind the current phasor.