Electromagnetic induction emf question

[EF17xx]

Homework Statement

The correct answer is D

Homework Equations

The Attempt at a Solution

The current is clockwise( in the first picture at least) due to right hand rule.

The magnetic flux is changing, the direction of the induced emf depends on the direction of change (increase or decrease)

Please help me solve this problem.

 

[Dr Dr news]

How does decreasing the resistance in the circuit above decrease the current?

 

[EF17xx]

How does decreasing the resistance in the circuit above decrease the current?

Decreasing the resistance means current can flow more easily, therefore ALL current chooses path of least resistance (through the copper wire) and so none goes through light bulb 2 ?

 

[Dr Dr news]

I agree that no current goes through L2. The question is why does L1 get dimmer?

 

[EF17xx]

I agree that no current goes through L2. The question is why does L1 get dimmer?

The area got smaller (this means that flux got smaller) from the equation flux = BAcos(theta)
When magnetic flux changes there is an emf induced, with a current that will oppose the change. Since flux got smaller we want it to increase right?
To increase it the direction of the induced current must be so that its magnetic field is in the same direction as the existing magnetic field to get more of it and increase the flux. When it its in the same direction the current is clockwise? as it is initially This would increase the current, and therefore actually make the lightbulb brighter.
Therefore I'm obviously wrong, what part of my thinking is wrong?

 

[Dr Dr news]

When I look at your diagram and read the problem, there is no reason to believe that dφ/dt is different after shorting out L2. It appears the the emf generated in the circuit is the same. The only difference is that the resistance has been halved. Where did answer D come from?

 

[EF17xx]

When I look at your diagram and read the problem, there is no reason to believe that dφ/dt is different after shorting out L2. It appears the the emf generated in the circuit is the same. The only difference is that the resistance has been halved. Where did answer D come from?

Yes ok I agree, but if the resistance is halved shouldn't the brightness increase? The answer came from the back of the book.

The problem states that the magnetic field is changing however wouldn't that mean that the current keeps changing direction also. Even though that still doesn't seem helpful to me because we don't know what the change is therefore don't know the current direction.

 

[Dr Dr news]

I'm afraid that finding a wrong answer to an end-of-chapter problem is not that rare. Although it hurts me to admit this, I too have made several mistakes this year so far and it's early.

 

[EF17xx]

I'm afraid that finding a wrong answer to an end-of-chapter problem is not that rare. Although it hurts me to admit this, I too have made several mistakes this year so far and it's early.

Thanks for your help and honesty nonetheless.

 

[cabraham]

Answer D could be right. We don't know the frequency of the mag field B. The loop has inductive reactance X & the resistance of the bulbs R+R. If X > 2R, then shorti g out o e lamp results in nearly the same current. The impedance of the loop goes from 2R + jX, then to R + jX. Since jX dominates, current hardly changes, increasing slightly.
But the look are decreases, which decreases enclosed flux. So the open circuit voltage & loop current decrease as well.
D is correct here.
But if X < R, then current may not decrease when lamp gets shorted. When X is small the loop impedance is dominated by 2R. Loop current is open circuit voltage divided by 2R. If lamp 2 gets shorted, then loop impedance is one times R plus jX. Since R > X, R dominates. Open circuit voltage drops due to decreased area, but impedance drops from 2R to R, neglecting jX as it is small. Smaller voltage with smaller impedance could result in current increasing, decreasing, or near the same.
Ok?

Claude