How to find the points between which EMF is induced in a loop with resistors?

Question

I recently came across a problem on Faraday's law, which had multiple resistors. The circuit looked something like this, with the loop rotating clockwise.

Personally, I felt that there would be three EMFs induced: one across the semi-circular loop PQ and another two across the quarter-circle areas. However, my book's equivalent circuit looks like this.

Why aren't we considering the EMF across the semicircle PQ?

PQ is a semi-circle that has no resistors in it. The rod PQ has a resistance of R and B exists everywhere, into the screen,

I think you need to start thinking about an induced voltage being in series with a loop and not across it. I have no idea what semicircle AB is. — Andy aka, Oct 18 '22 at 13:04
and also, isn't an EMF induced across a loop? At least that's what I was taught. — math and physics forever, Oct 18 '22 at 13:11
No it's induced in series with the loop. You can see that from the solution. — Andy aka, Oct 18 '22 at 13:12
I see, so if it's induced in series with the loop, why won't there be any EMF induced along the loop AB? It will still be in series with the loop, will it not? — math and physics forever, Oct 18 '22 at 13:18
Does the loop cross the magnetic field B? The loop is running parallel to the lines of flux. So no voltage will be induced. — StainlessSteelRat, Oct 18 '22 at 14:36
The field is perpendicualr to the loop, and teh rodPQ has a resistance R — math and physics forever, Oct 18 '22 at 14:39
I still have no idea which of the two semicircles PQ is. You haven't defined the extent of the field either; does it just exist at one single point? — Andy aka, Oct 18 '22 at 15:47
As far as I am aware (based on what you have added and described), a coil rotating perpendicular to a uniform magnetic field will produce no induced emf. No lines of flux will be cut hence no induced voltage. — Andy aka, Oct 18 '22 at 16:36

Mohammad Ali · Accepted Answer · 2022-11-03T18:03:38.637

The induced EMF in your drawn circuit will be only due to the rotating "rod". The whole circle will not contribute to any induced EMF.

The induced EMF in a moving conductor in a magnetic field is given by: $$emf = \int{(\vec{u}\times\vec{B})\,.d\vec{l}}$$ where \$\vec{u}\$ is the velocity of the moving conductor and the integral is done over the conductor path.

As you see,

for the whole rotating circle, the cross product is done between the velocity which is tangent to the circle and the magnetic field which is normal to the plane. This yields a component in the radial direction and normal to the path length l, which makes zero dot product after that.
but for the rod, the integral is done along the radial direction itself, and then the dot product is not zero.

This is summarized intuitively by noting that the perimeter of the circle is not crossing the magnetic field during the rotation process, only the rod PQ crosses the field.

Due to the symmetry, the induced voltage in the two rod halves will be equal to each other, and opposite in direction as the solution continues.

Another final point, your textbook does NOT consider an induced EMF in the resistor \$(2r)\$. Why then? Isn't it also crossing the field \$\vec{B}\$ ? The answer is; putting it in that schematic representation indicates that it is an ideal resistor, which means it has negligible physical dimensions. There is no physical length l to integrate the above cross product like the rod case.

How to find the points between which EMF is induced in a loop with resistors?

1 Answers1