I was given the following question:-

The plane figures shown are located in a uniform magnetic field directed away from the reader and diminishing. The direction of current induced in the loops is shown in the figure. Which one is the correct choice?

enter image description here

The correct answer is (D).

I am aware of Lenz’s Law, and I know how to find the direction of current given a changing magnetic field, and a closed circular loop. But this question is completely new to me.

Given a random circuit, and a changing magnetic field, is there a standard technique to predict the direction of current?

  • 1
    $\begingroup$ I'm assuming the part of the question that read "field directed away from the leader..." should be "field directed away from the reader..." i.e. into the page. $\endgroup$ – M. Enns Jun 6 at 13:26

Lenz's law tells you that the induced current in the loop must give rise to flux through the loop that is away from the reader. Then you can use the right hand grip rule to find the direction of current that gives rise to such flux.

I expect that this is what you did. Is it a standard technique? Probably.

Equivalently, you can set up the sign convention that an anticlockwise circulation as seen on paper (either emf or current) is given a positive sign, and flux towards the reader is given a positive sign. [This embodies a right handed association.] Then the emf and current follow from the Faraday-Lenz law, $$\mathscr E=-\frac {d \Phi}{dt}$$

For example, in your case, $\Phi$ is getting less negative so $\frac {d \Phi}{dt}$ is positive, so $\mathscr E$ is negative, that is clockwise.

Now let's look at the more complicated configurations in A, B and C. We know from Lenz's law that the induced current must produce flux into the page.

In A, we can deduce from symmetry that no current will flow in the diametric conductor. But the current in the outer loop would produce flux out of the page.

In B, the flux produced by the current shown in the outer loop would produce flux through itself out of the page. This flux is greater in magnitude than the flux produced into the page by the inner loop. [This is because, for a wire of radius, a, much less than then loop radius, r, the flux through the loop produced by a given current more than doubles when the loop radius doubles. In fact $\Phi=\mu_0 Ir \ln \left(\frac {1.39 r}{a}\right)$ to a good approximation.] So B is wrong.

In C the radial link might as well not be there. The loops are independent and will both have clockwise currents induced in them. So C is wrong.

I hope that a few more stack exchangers will take an interest in this question.

  • $\begingroup$ I think the wording of the question was poor. I was unable to solve this question , and I don’t know why the answer is (D) , as I’m only used to simple circular loops. $\endgroup$ – Aspirant Jun 6 at 13:02
  • $\begingroup$ A is in my opinion, the trickiest. I dealt with it by realising that, by symmetry, no current would flow in the diametric conductor. Therefore you need consider only the outer loop. In B consider the flux in the area bounded by the current, that is roughly the area between the two circular arcs. The currents shown both give flux out of the page (towards you) through that area. In C the single radial linkage might as well not be there. Although the flux between the loops is into the page, the flux through the inner loop is out of the page (as the field from the outer loop is the weaker here). $\endgroup$ – Philip Wood Jun 6 at 13:25
  • $\begingroup$ Maybe the question is incomplete? It seems to me that in cases (B) and (C) , the overall direction of flux depends on the dimensions of the loops... $\endgroup$ – Aspirant Jun 6 at 15:19
  • $\begingroup$ I have to apologise. I'd forgotten that the loops in C were independent. I've altered my answer – for the last time, I hope. $\endgroup$ – Philip Wood Jun 7 at 7:54
  • $\begingroup$ Everything is clear now. Thank you for the help. $\endgroup$ – Aspirant Jun 7 at 11:16

Well electromagnetism says three things;

1)relative motion between a magnet and a closed loop(circuit) creates current

2)orientation of magnet moving relative to a closed loop affects the direction of current

3)how fast the magnet is moving relative to a closed loop affects the rate of creation of cirrent

  • $\begingroup$ What I don't like about this answer (besides being very general and broad) is that it completely disregards induction by changing magnetic fields (which don't require a moving magnet at all) $\endgroup$ – Quantumwhisp Jun 6 at 16:40
  • $\begingroup$ Yes true a changing magnetic field does create a changing electric field. $\endgroup$ – user233873 Jun 6 at 17:01
  • $\begingroup$ If the magnetic field is changing at a constant rate, the associated electric field will be constant. Feel free to regard this as a nit-pick. $\endgroup$ – Philip Wood Jun 7 at 8:49
  • $\begingroup$ Yes thank you a lot. $\endgroup$ – user233873 Jun 7 at 9:00

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