Consider a small magnet that is suspended just above a conducting fixed loop placed. The magnet is oriented perpendicular to the axis of the loop and is released to fall under gravity and at the same instant current is allowed to flow in the loop.

Let me paint the picture here:
The x-y plane contains the loop of equation $x^2+y^2=1$. Gravity acts in the negative $k$ direction. The magnet is above the x-y plane. The north of the magnet points along positive x. While south points along negative x. The current in the ring is clockwise as seen from the top.

Note- The above equation is only to clarify the question and an answer involving these equations is not required.

My question is how will the magnet behave?

The current in the loop creates magnetic field in the negative z direction. As the magnet passes the loop, it must interact with the field of the loop. I don't think lenz law will help here as the loop already has current in it. The orientation of the magnet makes it hard to follow the situation. Maybe the maxwell equations can help?

My teacher says the magnet would rotate about the axis of the ring. I don't seem to follow the reason. All help will be appreciated. I will be happy to edit the question for further clarification if required.

  • $\begingroup$ > "at the same instant current is allowed to flow in the loop." Is this current due to a battery, or is the induced current due to motion of the magnet? In any case, the loop with electric current behaves as a magnet too, so its action on the falling magnet is similar to that of a magnet. $\endgroup$ Apr 2, 2018 at 14:56
  • $\begingroup$ @JánLalinský The current in the loop is from a battery. The magnet is perpendicular to the axis of the loop. $\endgroup$ Apr 3, 2018 at 9:45

1 Answer 1


Let's represent your magnet as a dipole, and we'll start with the simple case of no current in the ring. In that case the situation looks like this:

No current

As the magnet falls through the ring the net flux is zero because the north and south poles cancel each other out. That means no current is induced in the ring and the magnet falls straight through it accelerating at $g$.

Now turn on a clockwise current in the ring (I assume this is clockwise when viewed in the direction the magnet is falling). As you say this creates a dipolar field in the ring pointing downwards:

With current

The magnet rotates simply because the south pole of the ring field repels the south pole of the magnet and attracts the north pole of the magnet. How much the magnet rotates will depend on the magnitude of the current in the ring (i.e. the strength of the ring field) and how fast the magnet is travelling as it passes through the ring.

Once the magnet has rotated out of the x-y plane this breaks the symmetry and it means the magnet will now induce a current in the ring. As the magnet approaches the ring the current will be in the opposite direction to the current you are passing through the ring, then after the magnet has fallen through the ring and is moving away from it the induced current will be in the same direction as the current in the ring. The effect will be to slow the magnets fall.

How big the effect is will be a complicated calculation. It depends on how much the magnet rotates, which in turn depends on how large the current in the ring is and how fast the magnet is falling. It also depends on the exact shape of the magnets field - real magnets are never perfect dipoles.

  • $\begingroup$ +1 I must know, Where do you get so beautiful diagrams? $\endgroup$ Apr 30, 2018 at 15:57

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.