0
$\begingroup$

Suppose we have a loop of copper wire moving perpendicularly through a constant finite rectangular magnetic field directed into the screen .

When the loop enters the field, the induced current would be counter-clockwise and when it leaves the field the induced current would be clockwise according to Lenz's law if we view the loop moving through the field from above.

But the question here is that when the loop is moving through the middle of the field where it is neither entering nor leaving the field, can we say that the magnetic flux is changing and thus producing an induced current in the loop? The field is constant throughout, so will the change in flux be zero? But if not, how can we tell in which direction the current will be flowing?

How could Lenz's law be applied to such a case?

diagram

$\endgroup$
1

2 Answers 2

0
$\begingroup$

Since B and the area A are constant the flux B*A is constant so no current is induced, so there is no direction

$\endgroup$
4
  • $\begingroup$ there is no flux change in a homopolar dynamo either, still, it works $\endgroup$
    – hyportnex
    Commented Sep 22, 2023 at 1:23
  • $\begingroup$ @hyportnex different parts of a homopolar generator including the external circuit travel at different speeds. $\endgroup$
    – Farcher
    Commented Sep 22, 2023 at 6:54
  • $\begingroup$ @Farcher exactly and that crucial distinction is missing from the above explanation to be complete. $\endgroup$
    – hyportnex
    Commented Sep 22, 2023 at 12:43
  • $\begingroup$ @hyportnex there is change of flux if one considers the complete system $\endgroup$
    – trula
    Commented Sep 22, 2023 at 13:43
0
$\begingroup$

Faraday's law is applicable only for a stationary loop in space. You can't move the loop to a different position and say that the flux through the previous loop has changed. Here as the magnetic field is not varying with time no emf will be induced in the loop.

But when the loop is moving into the magnetic field region, initially only one of the sides is inside the region. As a result a motional emf is created across the side.

When the loop is leaving the region, the other side(opposite) will be the one inside the region and once again a motional emf is created across the side. So you end up getting that:

•When the loop is entering the region, current is anticlockwise.

•When the loop is leaving the region, current is clockwise.

•When the loop is completely inside the region, there is no current in the loop.

$\endgroup$

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