- #1

zenterix

- 530

- 74

- 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?

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