Brightness of two lamps in a circuit having a changing magnetic field

[songoku]

Homework Statement

Please see below

Relevant Equations

Faraday's law
Lenz's law

The answer key is (D) but I don't understand how to approach this question.

I am guessing the wire is acting as short circuit path but how to know which bulb will be short - circuited?

Thanks

 

[TSny]

I don't think D is the correct answer.

Instead of worrying about which bulb is "shorted", think about how you can use Faraday's law. For example, how would you use Faraday's law to determine the current in the light bulbs for the circuit on the left?

 

[songoku]

I don't think D is the correct answer.

Instead of worrying about which bulb is "shorted", think about how you can use Faraday's law. For example, how would you use Faraday's law to determine the current in the light bulbs for the circuit on the left?

It depends on the change of the magnetic field.

If the B increases then the current is anticlockwise and if the B decreases then the current is clockwise.

Thanks

 

[TSny]

It depends on the change of the magnetic field.

If the B increases then the current is anticlockwise and if the B decreases then the current is clockwise.

Yes. But we are interested in the brightness of the lamps. So, we are interested in the magnitude of the current in the lamps. For the single-loop circuit on the left, can you find an expression for the magnitude of the induced current in the loop, ##I##, in terms of the rate of change of the magnetic flux through the loop , ## \large \frac {d \Phi}{dt}##, and the resistance ##R## of each lamp?

 

[songoku]

Yes. But we are interested in the brightness of the lamps. So, we are interested in the magnitude of the current in the lamps. For the single-loop circuit on the left, can you find an expression for the magnitude of the induced current in the loop, ##I##, in terms of the rate of change of the magnetic flux through the loop , ## \large \frac {d \Phi}{dt}##, and the resistance ##R## of each lamp?

$$I=\frac{\frac{d\Phi}{dt}}{2R}$$

 

[TSny]

$$I=\frac{\frac{d\Phi}{dt}}{2R}$$

Yes.

Can you do anything similar in the multiloop circuit on the right?

 

[songoku]

Yes.

Can you do anything similar in the multiloop circuit on the right?

Assuming the resistance of the wire is negligible, the current flowing in the top loop will be bigger than the lower loop since the area is bigger so the rate of change of magnetic flux is also higher.

The current flowing in the top loop will be ##\frac{\frac{d\Phi_1}{dt}}{R}## and the current flowing in the bottom loop will be ##\frac{\frac{d\Phi_2}{dt}}{R}##.

 

[TSny]

Is there any magnetic flux through the bottom loop?

 

[songoku]

Is there any magnetic flux through the bottom loop?

Oh I see what it means of the picture. I just assumed that all part inside the loop had magnetic field.

So the induced current on the bottom is zero, that's why L2 goes out. For top loop, the decrease in area does not affect the rate of change of magnetic flux. The induced current will increase since the resistance of the circuit decreases

I suppose the appropriate answer will be (C).

 

[TSny]

Oh I see what it means of the picture. I just assumed that all part inside the loop had magnetic field.

Yes, I believe all of the flux that was in the first circuit is in the top loop of the second circuit and there is no flux though the bottom loop of the second circuit.

So the induced current on the bottom is zero, that's why L2 goes out.

Yes, there is no induced emf in the bottom loop. So, there can't be any IR voltage drop across the lower lamp.

For top loop, the decrease in area does not affect the rate of change of magnetic flux. The induced current will increase since the resistance of the circuit decreases

I suppose the appropriate answer will be (C).

Yes.

 

[songoku]

Thank you very much TSny