Emf induced in a conducting ring.

  • Thread starter srijag
  • Start date
  • #1
15
2
Consider a magnetic field perpendicular to a conducting ring moving with a velocity, v.When the ring is moving on the ground in translational motion alone, will emf be induced? I am slightly confused because if you consider the two halves of the ring as two rods, emf will be induced in both of them individually, but what about the emf induced in the ring as a whole?
 
Last edited:

Answers and Replies

  • #2
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,529
110
You have not given a clear picture of the relative orientations and motion, but if the magnetic flux through the ring changes with time, there will be an EMF.
 
  • #3
15
2
The ring is performing translational motion and the magnetic field is orthogonal to the motion of the ring. Does it matter if the magnetic flux is changing because I considered the two halves of the ring to be separate rods. That way emf induced in each 'rod' is E= Bl(2R) where R is the radius of the sphere.
 
  • #4
Nugatory
Mentor
12,988
5,699
The ring is performing translational motion and the magnetic field is orthogonal to the motion of the ring. Does it matter if the magnetic flux is changing because I considered the two halves of the ring to be separate rods. That way emf induced in each 'rod' is E= Bl(2R) where R is the radius of the sphere.
Consider the direction of the EMF induced in the two separate rods.

Suppose we were to connect the left ends of the two rods, and the right ends of the rods, with conducting wires. That's equivalent to the ring. What's the induced EMF across the two connections?
 
  • #5
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,529
110
Two rods are not a ring. However you try, the EMF is given by the rate of change of the magnetic flux through the ring.
 
  • #6
15
2
Consider the direction of the EMF induced in the two separate rods.

Suppose we were to connect the left ends of the two rods, and the right ends of the rods, with conducting wires. That's equivalent to the ring. What's the induced EMF across the two connections?
If you apply Kirchoff's loop rule over here, the total EMF will be zero.
 
  • #7
Philip Wood
Gold Member
1,221
78
MA: I agree that looking at the change (if there is any!) in flux through the ring is the best approach. Nugatory is (I think) trying to come at it from the OP's point of view of regarding the ring as joined rods. The emf in each will be in the same direction in space, but looking at the rods as parts of the ring, in opposite senses around the ring.
 
  • #8
15
2
Two rods are not a ring. However you try, the EMF is given by the rate of change of the magnetic flux through the ring.
Nothing is specified about the magnetic field. So it might also be a constant field which means there won't be any EMF induced.
 
  • #9
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,529
110
Nothing is specified about the magnetic field. So it might also be a constant field which means there won't be any EMF induced.
If it's a constant B field, there's no point asking the question.
 
  • #10
Philip Wood
Gold Member
1,221
78
With respect, MA, there might be a point, if the OP is trying to reconcile emf's induced in conductors cutting flux (the 2 rods approach) with there being no emf in the ring.
 
  • #11
15
2
For the emf to be induced in a moving conductor, you don't require varying magnetic field. So, that's the reason i considered them to be two rods. In that case, emf will be produced in both as in the case of two cells connected in the same circuit. I need help with that part of the problem.
 
  • #12
WannabeNewton
Science Advisor
5,800
532
MA: I agree that looking at the change (if there is any!) in flux through the ring is the best approach. Nugatory is (I think) trying to come at it from the OP's point of view of regarding the ring as joined rods. The emf in each will be in the same direction in space, but looking at the rods as parts of the ring, in opposite senses around the ring.
A ring is not the same thing as two rods, no matter how you try to construe the system.

OP, why not just start with two conducting rods connected by wires like Nugatory suggested?

Now, why does EMF occur in the first place for a conducting system moving non-trivially through an external magnetic field? What does the magnetic field do to the charges in the conducting system that causes an EMF to be induced? And how does this relate to the "two conducting rods connected by wires" system?
 
Last edited:
  • #13
Philip Wood
Gold Member
1,221
78
A ring is not the same thing as two rods, no matter how you try to construe the system.
I did say "joined rods", imagining their ends to be connected together to form a ring.
 
  • #14
15
2
Now, why does EMF occur in the first place for a conducting system moving non-trivially through an external magnetic field? What does the magnetic field do to the charges in the conducting system that causes an EMF to be induced? And how does this relate to the "two conducting rods connected by wires" system?
When a conducting body moves through the field, the charges in the system experience force 'qvb' and hence, opposite charges move to extreme ends and potential difference is established. this continues until qvb= qe(i.e; force due to electric field in the direction opposite to that of qvb.)
 
  • #15
Philip Wood
Gold Member
1,221
78
srijag. Do you now have a satisfactory answer to your original question?
 

Related Threads on Emf induced in a conducting ring.

Replies
23
Views
5K
Replies
3
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
28
Views
6K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
2
Views
8K
  • Last Post
3
Replies
59
Views
10K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
3K
Top