Induced emf

  • Thread starter Pixter
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  • #1
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If we have a loop with radius 12 cm that lies in a uniform magnetic field of 1.5T.

The magnetic field is in the positive z direction and the loop lies in the horizontal xy-plane. Now if the loop is removed from the field in a time interval of 2.0x10^-3 s what will the average induced emf be.



Right now first I thaught it might be zero but then seeing as the b field will be constant and A will change.

therefore it has to be emf= - -b x dA/dt

but how can i describe the rate of change of A?


ps:am I even on the right track? =)
 
Last edited:

Answers and Replies

  • #2
PPonte
If I understood well the magnetic field is perpendicular to the loop:
the A will decrease from it's maximum value (the area of the loop: 4,52 x 10^-2 m^2) to zero.
 
  • #3
30
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PPonte said:
If I understood well the magnetic field is perpendicular to the loop:
the A will decrease from it's maximum value (the area of the loop: 4,52 x 10^-2 m^2) to zero.
yeah so the average emf induced would be d(A/2)/dt ?
 
  • #4
PPonte
In my opinion, we could simplify the formula:
emf induced = B.A/dt

B= 1,5 T
A (of the loop) = 4,52 x 10^-2 m^2
dt = 2,0 x 10^-3 s
emf = 33,9 V

But wait for the experts, it's possible that I am mistaken.
 
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  • #5
79
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I thinks it is

[tex] \frac{\Delta \Phi_B}{\Delta t} [/tex]

since it asks average value not instantanious one, I think 33.912 V is a correct answer.
 
  • #6
30
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Psi-String said:
I thinks it is

[tex] \frac{\Delta \Phi_B}{\Delta t} [/tex]

since it asks average value not instantanious one, I think 33.912 V is a correct answer.
but surely the average A confined in the magentic field during dt must be half A.. therefore the average Induced emf must be -d(A/2)/dt

or?
 
  • #7
PPonte
It is not an average A but a variation of A.
Phi=B.A.costheta

and

efm = dPHi/dt

PS- Sorry, but I don't know to latex.
 
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