Induction and EMF

  • #1

Homework Statement


In a ring of radius R, there is current I flowing in the counterclockwise direction as viewed from above. A small ring of radius r is on a common axis and is a hight z above the current carrying ring, where z>>R. The small ring moves up with velocity v. Calculate the emf in the upper ring and the direction of the induced current. It can be assumed that the ring is sufficiently small so that the magnetic field across it's area is constant.


Homework Equations





The Attempt at a Solution


So E= - dΦ/dt
Φ = B*A ( since the magnetic field across it's area is constant ).
A = π*r*r
B= (μIR^2) / [ 2 ( R*R + z*z )^3/2 ] but since z>>R we can use the approximation
B= (μIR^2) / (2z^3)

Now i know I need to do the -dΦ/dt but I am getting confused ( how to do it? ).
Also the direction of the induced current will be clockwise. Please help
 

Answers and Replies

  • #2
G01
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In order to do the derivative you need to have time involved. The only quantity in that expression for the flux that depends on time is z, correct? Is there any way to get z in terms of time?

HINT: Have you used the velocity, v, yet?
 
  • #3
So will the emf be A* (μIR*R/2) * 3( v^3 * z^4 ) or I did it very very wrongly..
 
  • #4
G01
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You should not have z in your final answer. Start with this line:

[tex]\Phi=\frac{\mu IR^2A}{2z^3}[/tex]

Then, remember that: v = z/t.

Can you use the last expression to remove z from the first?
 
  • #5
Sorry I made a typo.. should have been A* (μIR*R/2) * 3/( v^3 * t^4 ) but then still
A* (μIR*R/2) * 3/( z^3*t )
 
  • #6
Is this anywhere near to correct?
 
  • #7
G01
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I think you are missing a minus sign in there. You should have picked up a -3 with the differentiation.

Other than that, it looks good.
 
  • #8
But from the original equation which is emf = - dΦ/dt, so this minus cancels out with the -3 minus.. I will redo it just in case. That's for all the help, you can count this one as solved ( I understood the main idea with the z being the only thing that changes with time! ). Thanks very much
 

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