Understanding self-inductance in a solenoid.

[zenterix]

Homework Statement

I'd like to understand the concepts of mutual and self inductance.

Relevant Equations

Suppose we have the following setup

That is, a solenoid with $N$ turns, length $l$, radius $R$, and a current $I$ flowing.

Let's approximate the solenoid as an infinite solenoid (ie, $l$ is very large).

Then, the magnetic field inside the solenoid is

$$\vec{B}=\mu_0 nI\hat{k}=\frac{\mu_0NI}{l}\hat{k}\tag{1}$$

Suppose $\frac{dI}{dt}> 0$. Then, in the solenoid the magnetic field is changing and so is the magnetic flux.

This variance of the magnetic field creates a phenomenon called electromagnetic induction.
Faraday conjectured that this phenomenon is due to a non-electrostatic electric field $\vec{E}$ which satisfies

$$\mathcal{E}=\oint_C\vec{E}\cdot d\vec{s}=-\frac{d\Phi_B}{dt}=\frac{d}{dt}\iint\vec{B}\cdot d\vec{A}\tag{2}$$

where $\mathcal{E}$ is called the electromotive force. This isn't an actual force. It has units of voltage. It is like a potential difference but isn't one either since this new electric field isn't conservative.

As I understand it, in the case of the solenoid, we have $N$ turns, not loops.

On the other hand, we can approximate the solenoid as being composed of $N$ loops.

The flux through the solenoid is then just the sum of the flux through the $N$ loops.

$$\Phi=N\iint_{\text{turn}}\vec{B}\cdot d\vec{A}=n^2l\mu_0n\pi R^2I$$

(I am not sure about the term "flux through the solenoid" above. Flux is defined relative to a surface, so this term doesn't seem correct).

There is a linear relationship between $\Phi$ and $I$. The proportionality constant is the self-inductance $L=n^2l\mu_0n\pi R^2$.

Why is this important?

It seems to be because we have the concept of self-induced emf $\mathcal{E}_L$.

$$\mathcal{E}_L=-N\frac{d\Phi_B}{dt}\tag{3}$$

$$=-N\frac{d}{dt}\iint_{\text{turn}}\vec{B}\cdot d\vec{A}\tag{4}$$

$$=-N\frac{d}{dt}\left (\frac{\mu_0 NI}{l}\pi R^2\right)\tag{5}$$

$$=-\frac{N^2\mu_0\pi R^2}{l}\frac{dI}{dt}\tag{6}$$

$$=-n\mu_0l\pi R^2\frac{dI}{dt}\tag{7}$$

If this back emf were not present, the magnetic field would simply increase due to the presence of $I$ in (1).

At this point, as I understand it, we add the whole new field $\vec{E}$ that we mentioned previously. It doesn't come from previous theory, but rather from experiments.

This is the part I'd like to understand more deeply.

I have a few questions but let me tackle them in parts.

The negative sign in equations (3)-(7) are there because of Lenz's law, which seems to also be simply a law obtained from experimental evidence.

$$\mathcal{E}_L=-L\frac{dI}{dt}=-n^2l\mu_0 \pi R^2 \frac{dI}{dt}=\oint_C \vec{E}\cdot d\vec{s}\tag{8}$$

The rightmost integral has units of voltage of course, which is energy per charge.

But what is $C$ in the case of our solenoid?

 

[zenterix]

Let me move to my next doubts.

From (8), we have $\mathcal{E}_L<0$. Given our choice of normal vector $\hat{k}$ in the calculation of magnetic flux, it seems that $\vec{E}$ has fieldlines going clockwise in the solenoid wires.

There is an induced current opposite to the initial current which "generates" a field that opposes the increase in magnetic field. This induced magnetic field thus has field lines pointing down inside the solenoid.

Where in the calculations do we see this induced current?

We assumed, after all, that $\frac{dI}{dt}>0$. But if there is induced current, isn't this rate of change affected?

 

[zenterix]

I'm going to try to guess my way through the questions I posed.

I'm guessing that since we approximated the solenoid as $N$ loops, then $C$ is just one of the loops.

$$\mathcal{E}_L=-N\frac{d\Phi}{dt}=\oint_C\vec{E}\cdot d\vec{s}\tag{9}$$

$$\frac{\mathcal{E}_L}{N}=-\frac{d\Phi}{dt}=\frac{1}{N}\oint_C\vec{E}\cdot d\vec{s}\tag{10}$$

Does this mean that the back emf on one loop is $\frac{1}{N}$ the back emf on the entire solenoid?

Now, this division by $N$ doesn't seem to matter because

$$\frac{\mathcal{E}_L}{N}=\frac{1}{N}\oint_C\vec{E}\cdot d\vec{s}\tag{11}$$

$$\implies \mathcal{E}_L=\oint_C\vec{E} \cdot d\vec{s}\tag{12}$$

By symmetry, $\vec{E}$ is constant on a loop.

$$\mathcal{E}_L=E2\pi R\tag{13}$$

$$E=\frac{\mathcal{E}_L}{2\pi R}\tag{14}$$

$$=-L\frac{dI}{dt}\frac{1}{2\pi R}\tag{15}$$

Now, to be honest, I am not at all happy with this post as it seems very fishy and incorrect.

Is $C$ one of the loops? At this point I really don't know.

 

[kuruman]

Closed loop $C$ is anything you choose it to be and encloses area $S$ over which the flux is changing. See my post #2 in your earlier post. If you choose $S$ to be the area of one turn, then $C$ is the circumference of one loop and $\mathcal E$ is the emf you get across an infinitesimally small cut in the loop. If you choose $S$ to be the area of all the turns, then $C$ is the circumference of all the truns and $\mathcal E$ is the emf you get across the two ends of the solenoid.