This question is motivated by Problem 7.12 in Griffiths Electrodynamics book. I have already solved the question, so there is no issue with homework policy.

Relevant Equations: $$\Phi = \int{\vec{B}\cdot d\vec{a}}$$ $$\mathcal{E} = -\frac{d\Phi}{dt}$$ $$I = \mathcal{E}/R$$

The problem states: A long solenoid, of radus a, is driven by an alternating current, so that the field inside is $$\vec{B}(t)= B_0\cos(wt)\hat{z}$$ A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time.

My solution was to take the field of the solenoid, integrate it over the loop, take the time derivative and use the relation $$\mathcal{E} = -\frac{d\Phi}{dt}$$ Then, one simply takes $$I = \frac{\mathcal{E}}{R}$$ This works just fine, and you'll get the correct answer: $$\frac{\pi a^2wB_0\sin(wt)}{4R}$$

However, my concern is that we are ignoring the self inductance. Because the current in the loop is changing, this also creates an emf. The equation we should start with is $$\Phi = \int{\vec{B}\cdot d\vec{a}}$$ taken over the loop's surface. But, the field is no longer the field due simply to the solenoid, but the field should include the contribution from the loop itself because it now has some current running that is changing in time. This would be enormously more complicated as it's hard to calculate the field off axis of a loop. But, shouldn't we include this in the flux calculation?


1 Answer 1


But, shouldn't we include this in the flux calculation?

You can think about this problem like coupled inductors where the 'voltage across' the 'secondary' (inner loop of wire) is given by

$$v_s = k\sqrt{L_1L_2}\frac{di_1}{dt} + L_2\frac{di_2}{dt}$$

where $L_1$ and $L_2$ are the self-inductance of the solenoid and inner loop respectively, and $0\lt k \lt 1$ is the coupling coefficient.

So yes, there is a contribution due to (rate of change) of the current through the inner loop but, under the plausible assumption that $k\approx 1$ and $L_1\ggg L_2$, it is the mutual inductance term that dominates.

There is a related example of calculating the mutual inductance of a few turns of wire around the outside of a solenoid on page 8 of the document here:

enter image description here

  • $\begingroup$ Thanks. I'm not completely convinced that L_1 >> L_2 is always that obvious, why do you think this is so plausible? $\endgroup$
    – user7348
    Sep 18, 2019 at 1:50
  • $\begingroup$ @user7348, Griffiths states that the solenoid is long which essentially means that $N_1$ is large while $N_2$ is given to be 1. Since the self-inductance goes as $N^2$, it's plausible here that $L_1\ggg L_2$. Note that I don't mean to imply that this is always the case, just that it is here. $\endgroup$ Sep 18, 2019 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.