Question about magnetic induction.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 1K views
Clara Chung
Messages
300
Reaction score
13

Homework Statement


Q.png


Homework Equations

The Attempt at a Solution


For cii), the answer is:
Yes. When the rod moves down, it moves across the horizontal field, cutting field lines. A current will be induced to oppose the motion. The magnetic force due to the motion is vertically upward.

But I think that, the voltage is induced in a direction perpendicular to the axis along PQ. No current can be formed. Therefore the rod will slide down as normal.

Why am I wrong? Thanks again.
 
Physics news on Phys.org
Label the bottom two corners A and B ... A is below P and B is below Q.
By your reasoning, could there be a voltage induced between A and P?

If there were a voltage induced between P and Q, wouldn't the same voltage also be between A and B ... after all, these are conductors?
 
  • Like
Likes   Reactions: Clara Chung
Simon Bridge said:
Label the bottom two corners A and B ... A is below P and B is below Q.
By your reasoning, could there be a voltage induced between A and P?

If there were a voltage induced between P and Q, wouldn't the same voltage also be between A and B ... after all, these are conductors?

Yes, a voltage can be induced between A and P. However the same voltage is induced between B and Q. The situation will be just like a circuit with two batteries oppose to each other. Shouldn't there be no current?
 
Simon Bridge said:
If there were a voltage induced between P and Q, wouldn't the same voltage also be between A and B ... after all, these are conductors?
I'm so sorry that I don't get your point, do you mean the situation in ci)?
 
When the magnetic field is vertical - you have no problem with there being a current along the rod between P and Q right?
Does that mean there is a voltage induced between P and Q?
 
  • Like
Likes   Reactions: Clara Chung
Simon Bridge said:
When the magnetic field is vertical - you have no problem with there being a current along the rod between P and Q right?
Does that mean there is a voltage induced between P and Q?
yes it does.
 
Sorry for asking stupid question. I think I now have problem with there being a current along the rod between P and Q. When a voltage is applied on PQ. That means PQAB is a short circuit. How do I account for the p.d. between AB?
By the way, there won't be induced emf on AB because it is not cutting any field lines.
 
Clara Chung said:
By the way, there won't be induced emf on AB because it is not cutting any field lines.
Correct.
Clara Chung said:
How do I account for the p.d. between AB?
That will depend on the resistances of the rod and the overall path, which are not given. The motional emf in the rod will be B*l*v, which will drive a current through the loop PQAB and ∫along loop PQABE.dl=dΦ/dt= B*l*v.
 
  • Like
Likes   Reactions: Clara Chung
cnh1995 said:
Correct.

That will depend on the resistances of the rod and the overall path, which are not given. The motional emf in the rod will be B*l*v, which will drive a current through the loop PQAB and ∫along loop PQABE.dl=dΦ/dt= B*l*v.

Thanks and please take a look at the questions on #1 and #3.
 
Clara Chung said:
But I think that, the voltage is induced in a direction perpendicular to the axis along PQ. No current can be formed. Therefore the rod will slide down as normal.
No. Apply the right hand rule and you'll see that an emf is induced in the rod along its length and it is equal to Blvsinθ, where θ is the angle between velocity vector and the magnetic field direction.
Clara Chung said:
Yes, a voltage can be induced between A and P. However the same voltage is induced between B and Q. The situation will be just like a circuit with two batteries oppose to each other. Shouldn't there be no current?
This would be true if the entire loop PQAB were moving. In this problem, since only the rod PQ is moving, motional emf is induced only in the rod, which circulates a current in the loop.
 
  • Like
Likes   Reactions: Clara Chung