## Induced electric fields

...an "Ampere's Law" for the electric field.

Faraday's law can be written as: $${\cal E} = \oint_{\cal P} \myv E \cdot d \myv l = -\frac{d}{dt}\int_{\cal S} \myv B \cdot d \myv a.$$

The path integral around a closed path can be written, according to Stoke's theorem, as a surface integral of a cross product, so: $$\int_{\cal S} \myv \grad \times \myv E \cdot d \myv a = -\frac{d}{dt}\int_{\cal S} \myv B \cdot d \myv a.$$

If the magnetic field is changing, but *not* the loop (and thus the surface), the differential can be passed under the integral like...
$$\int_{\cal S} \myv \grad \times \myv E \cdot d \myv a = -\int_{\cal S} \frac{\del \myv B}{\del t} \cdot d \myv a.$$

To be true for all fixed surfaces, it has to be the case that $$\myv \grad \times \myv E = - \frac{\del \myv B}{\del t}.$$

With magnetic fields, this was the sequence of things that lead to Ampere's law: $$\myv \grad \times \myv B = \mu_0 \myv J$$ $$ \text{and } \myv \grad \cdot \myv B = 0$$ $$\Rightarrow \oint \myv B \cdot d \myv l = \int \mu_0 \myv J\cdot d\myv a=\mu_0 I_\text{enc}.$$

Now with induced fields, as long as there's no charge density around, $\rho=0\propto\myv \grad \cdot \myv E$. So we can write... $$\myv \grad \times \myv E = -\frac{d}{dt}\myv B$$ $$\myv \grad \cdot \myv E = 0]$$ $$\Rightarrow \oint \myv E \cdot d \myv l =\int\(-\frac{d}{dt}\myv B\)\cdot d\myv a= -\frac{d}{dt} \Phi_B.$$

It looks like $-\frac{d}{dt} \Phi_B$ plays the same role as $I_\text{enc}$ played in magnetostatics. So we can use our whole, symmetry-related "bag of tricks" from Ampere's law now on electric fields where the change of flux plays the role of the enclosed current. For example...

**Example:** *An infinitely long straight wire carries a slowly varying current $I(t)$.
What is the induced electric field near the wire? First off...what direction
is the electric field?*

Try putting up Amperian loops where the surface area is perpendicular to... $\uv s$, $\uv z$, and $\uv{\phi}$ in turn... In which geometry is there a changing $B$-flux through the loop?

Convince yourself that it's only when the loop is perpendicular to $\uv{\phi}$
that there's a change in the flux.

Then, we've got: $$\oint \myv E \cdot d\myv{l} = E_z(s_0)l - E_z(s)l = -\frac{d}{dt}\int \myv B \cdot d \myv a,$$

eventually... $$\Rightarrow \myv E(s) = \left(\frac{\mu_0}{2\pi} \frac{dI}{dt} \ln s + C\right) \uv{z}.$$

### Inductance

A current $I_1$ flows in loop 1, giving rise to a magnetic field $\myv B_1$. What's the flux $\Phi_2$ through loop 2 of the field $\myv B_1$? $$\Phi_2 = \int \myv B_1 \cdot d\myv{a}_2.$$

We can in principle calculate $\myv B_1$ through the B-S law: $$\myv B_1 = \frac{\mu_0}{4 \pi}I_1 \oint \frac{d\myv{l}_1 \times \myv \rr}{\rr^2}.$$

Without actually solving this integral, we can see that $\myv B_1 \propto I_1$
which in turn implies $\Phi_2 \propto I_1$, that is..
$$\Phi_2 = M_{21}I_1,$$
where the constant of proportionality $M_{21}$ is called the **mutual inductance**.
(Units: "henries" (H), where 1 H = 1 volt-second / ampere).

Developing this further by means of the magnetic vector potential... $$\begineq \Phi_2 &=& \int \myv B_1 \cdot d\myv{a}_2 = \int \left(\myv \grad \times \myv A_1\right)\cdot d \myv a_2\\ &=& \oint \myv A_1 \cdot d\myv l_2.\endeq$$

We've seen the general expression for the magnetic vector potential: $$\myv A_1 = \frac{\mu_0 I_1}{4 \pi} \oint \frac{d \myv l_1}{\rr}.$$

Subbing... $$\begineq\Phi_2 &=& \frac{\mu_0 I_1}{4 \pi}\oint \left(\oint \frac{d\myv l_1}{\rr}\right)\cdot d\myv{l}_2\\ &=& \left[\frac{\mu_0}{4 \pi}\oint \oint \frac{d\myv{l}_1 \cdot d\myv{l}_2}{\rr}\right] I_1.\endeq$$

This is the **Neumann formula**. The quantity in square brackets $[..]=M_{21}=M_{12}\equiv M$.
Apparently the mutual inductance only depends on the geometric arrangement
of the current loops, and not the field.

Now, just as the flux in a second loop is proportional to the current in the
first loop, there is also a dependence of the flux through one loop, on the
current in that same one loop itself:
$$\Phi = L I,$$
where $L$ is the **self-inductance**.

An ${\cal E}$ (a 'back-${\cal E}$') will be induced by any change in the flux, which is (for rigid loops)... $${\cal E} = -\frac{d}{dt}\Phi = -L \frac{dI}{dt}.$$

We shall shortly see that $L$ acts like **mass** in an electrical circuit, lending the circuit some "inertia".

### L-R circuit

Every circuit, whether it has an inductor in it or not, is a current loop and so has some self inductance $L$. This self inductance is pictured as an additional circuit element here, but it's inherent in the circuit.

If current is flowing already, **unplugging** (or suddenly cutting) the circuit causes a very rapid $\frac{dI}{dt}$, and a spark is often generated.

What about when you **suddenly plug a circuit in**? $I(t=0) = 0$.

We'll have $$V_0 + {\cal E} = V_0-L\frac{dI(t)}{dt} = I(t)R.$$

To solve this, re-write it as... $$-\frac{L}{R}\frac{dI(t)}{dt} = I(t)-\frac{V_0}{R}.$$

Now define $F(t) \equiv I(t)-\frac{V_0}{R}$, and note that $\frac{dF}{dt} = \frac{dI}{dt}$, so we can re-write this equation as... $$-\frac{L}{R}\frac{dF(t)}{dt} = F(t).$$

The solution is an exponential... $$F(t) = ke^{(-R/L)t}=I(t)-V_0/R,$$

And so we conclude... $$\Rightarrow I(t)= ke^{(-R/L)t} + V_0 /R.$$

If $I(t=0)=0$, then $k=-V_0 /R$, so... $$I(t)=\frac{V_0}{R}(1-e^{(-R/L)t}).$$

## Magnetic energy

Wardenclyffe Tower was Nicola Tesla's attempt to transmit energy and messages across the Atlantic from Long Island to Europe.

Just as we found that there is an energy $W=\frac{1}{2}\int_{\cal V}\myv D\cdot \myv E\,d\tau$ associated with an electric, field, I want to show that there is also an energy that can be calculated from the magnetic field.

If some sort of voltage source (battery) $V_0$ is applied to a circuit, as above in general we have $$V_0 + {\cal E} = IR.$$

The work done by the battery in moving a small chunk of charge $dq=I\,dt$ once around the circuit is $$\begineq dW &=& V_0 dq = V_0Idt=-{\cal E}Idt+I^2Rdt\\ &=&Id\Phi + I^2R\,dt.\endeq$$

That last term is apparently Joule heating, and represents energy **irreversibly
lost** to heat.

The first term is work that is done to set up the magnetic fields, and can
be **reversibly extracted** from the system.

For a single, rigid loop circuit, $\Phi = LI \Rightarrow d\Phi = LdI$. So the work done to change the magnetic state of things $dW_B$ is $$dW_B=LI\,dI.$$

We can integrate this to get the total work done, starting from zero current... $$\begineq W_B &=& \int_0^ILI'dI' \\ &=& \frac{1}{2}LI^2 = \frac{1}{2}\Phi I=\frac{1}{2}\Phi^2 /L.\endeq$$

#### ..in terms of magnetic fields

$$\Phi = \int_{\cal S} \myv B\cdot d\myv a = \int_{\cal S} (\myv \grad \times \myv A)\cdot d\myv a = \oint_{\cal C} \myv A\cdot d\myv{l}.$$We can substitute this in to our work expression $$W_B=\frac{1}{2}\oint_{\cal C} I \myv A\cdot d\myv{l}.$$

It would be nice to write this in terms of current *density* instead
of total current. Substituting $I\,d\myv{l} \to \myv J\,d \tau$ and then $\oint_{\cal C}
\to \int_{\cal V}$...
$$W_B =\frac{1}{2} \int_{\cal V} \myv A \cdot \myv J d \tau =\frac{1}{2\mu_0} \int_{\cal V} \myv A \cdot (\myv \grad \times \myv B) d \tau.$$

Pulling out handy vector identity #6, we can re-write... $$\myv A\cdot (\myv \grad \times \myv B) = \myv B \cdot \myv B - \myv \grad \cdot (\myv A \times \myv B).$$

Using this in the energy expression:
$$\begineq W_B &=&\frac{1}{2\mu_0} \left[\int_{\cal V} B^2 d \tau -\int \myv \grad \cdot
(\myv A \times \myv B) d \tau\right]\\
&=& \frac{1}{2\mu_0} \left[\int_{\cal V} B^2 d \tau -\oint_{\cal S} (\myv A \times \myv B ) \cdot d\myv{a}\right].\endeq$$

Take ${\cal V}$ and ${\cal S}$ to be over all space--Why not, since we'll for sure catch the volume where the source currents are non-zero that way. Some things to note about the $\oint_{\cal S}$ integral

- $\myv A \propto 1/r,$
- $\myv B \propto 1/r^2,$
- $d \myv a \propto r^2.$

So, as $r\to \infty$, that second integral is dropping off as $1/r \to 0$. So we can neglect it compared to the first integral...

$$W_B = \frac{1}{2\mu_0}\int_{\cal V} B^2 d \tau.$$