Say we drop a current loop into a uniform magnetic field. As long as only on side of the loop is in the field, the magnetic flux through the loop is increasing and we hence have an induced voltage. The direction of the voltage is such, that the current it gives rise to results in a net force pointing upwards - working against the fall according to Lenz's law. This far this good.

However, what happens when the top end of the loop enters the field? By symmetry the current running in this branch will give a force pointing downwards, cancelling the one from the bottom. Hence, I believe the loop will be in simple free fall. But is there still a voltage induced?

The flux through the loop is not changing, but it is still "cutting lines of flux" which is often the argument used about Faraday's disk - a disk simply rotating in a uniform field and having an induced voltage. If there is a voltage induced in the falling loop too - how does that comply with energy conservation?

  • $\begingroup$ This question could be better understood if you improved the wording and if you gave some information on the (relative) orientation of the magnetic field of the loop and its movement. Also you are probably not dropping a "current loop" but a conducting loop. Otherwise you'd describe a loop of unchanging current moving around in a magnetic field. $\endgroup$ – freecharly Dec 29 '17 at 20:38

One way of thinking of the system is to consider the top nad bottom of the loop separately.

When the bottom of the loop is in the magnetic field an emf, $\mathcal E = BLv$, is induced and as there is a complete conducting circuit there is an induced current produced which is shown in the left hand diagram.
This induced current is producing a magnetic field out of the screen and trying to reduce the flux change producing it.

enter image description here

When the top part of the loop enters the magnetic field, as shown in the right hand diagram, it too produces an emf of the same magnitude and direction as that of the bottom loop but you will note that the induced current (as shown in red) is of the same magnitude but it is in the opposite direction to the induced current produced by the bottom part of the loop.
So the net current in the whole loop is zero as os the net induced emf as one would expect as there is no net change of magnetic flux linked with the loop.
Since there is no induced current there is no opposition to the motion of the loop.

In terms of energy changes in the first case there is a current flowing through the loop and so ohmic heating occurs which comes at the expense of a loss of either kinetic energy or (gravitational) potential energy of the loop whereas in the second case there is no such transformation of energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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