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EMF and magnetic flux

  • Thread starter gramentz
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  • #1
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Here is how the question is stated exactly:

A circular conducting loop is held fixed in a uniform magnetic field thatt varies in time according to B(t) = B0exp(-at) where t is in s, a is in s-1 and B is the field strength in T at t=0. At t=0, the emf induced in the loop is 0.0758 V. At t= 1.76 s, the emf is 0.0192 V. Find a. Answer in units of s-1



Homework Equations


emf= [tex]\frac{d\Phi}{dt}[/tex]


The Attempt at a Solution



I'm quite lost on this problem! I tried plugging in 0.0192 = B(t)/ 1.76 and solve for a, but obviously that did not produce the correct answer. Any help would be appreciated!
 

Answers and Replies

  • #2
gabbagabbahey
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Hint: How is magnetic flux related to magnetic field?
 
  • #3
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The magnetic flux is a way of counting the total magnetic field lines. A changing magnetic field (given by the B(t)) will produce the emf's given.
 
  • #4
gabbagabbahey
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The magnetic flux is a way of counting the total magnetic field lines.
No it isn't (thinking of it in this way allows you to visualize things, but the total number of field lines is infinite for all non-zero fields)

There is an equation that tells you how to determine the magnetic flux through a surface from the magnetic field (it involves an integral)...what is that equation?
 
  • #5
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[tex]\int[/tex]BdA.......B(dot)dA = BAcos[tex]\theta[/tex]
 
  • #6
gabbagabbahey
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Right, when the magnetic field is uniform (such as in this case), the flux is given by [itex]\Phi=\textbf{B}\cdot\textbf{A}=BA\cos\theta[/itex]

What is the surface area [itex]A[/itex] in this case? What does that tell you the EMF is?
 
  • #7
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I am not sure I understand because I don't see any values given in the problem for the dimensions of this loop. Do you mean A = [tex]\frac{\phi}{B}[/tex]?
 
  • #8
gabbagabbahey
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Sorry, you aren't given [itex]A[/itex], and you don't need to find it.

You have two equations, [itex]\text{EMF}=-\frac{d\Phi}{dt}[/itex] and [itex]\Phi=BA\cos\theta[/itex]...put them together...what do you get?
 
  • #9
gabbagabbahey
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Sorry, you aren't given [itex]A[/itex], and you don't need to find it.

You have two equations, [itex]\text{EMF}=-\frac{d\Phi}{dt}[/itex] and [itex]\Phi=BA\cos\theta[/itex]...put them together...what do you get?
 
  • #10
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EMF= -[tex]\frac{B_{0}exp(-at)A}{dt}[/tex] ? Or take the derivative of BAcos[tex]\phi[/tex] with respect to time and set that equal to EMF? (Thank you for your help thus far though)
 
  • #11
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I'm thinking I have to use the emf at time t=0 to find the initial B at time t=0.
 
  • #12
gabbagabbahey
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EMF= -[tex]\frac{B_{0}exp(-at)A}{dt}[/tex] ? Or take the derivative of BAcos[tex]\phi[/tex] with respect to time and set that equal to EMF? (Thank you for your help thus far though)
When you substitute your equation for [itex]\Phi[/itex] into the emf equation, you get

[tex]\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)[/tex]

Do [itex]A[/itex], [itex]\cos\theta[/itex] or [itex]B_0[/itex] depend on time? Why or why not?

What does that make the derivative?
 
  • #13
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The only one that would be dependent on time would be B[tex]_{0}[/tex] because the loop is held fixed and the angle does not change.
 
  • #14
gabbagabbahey
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[itex]B_0[/itex] is a constant. The only thing that varies with time is the exponential, so

[tex]\text{EMF}=-\frac{d}{dt}\left(B_0e^{-at}A\cos\theta\right)=-B_0A\cos\theta\frac{d}{dt}\left(e^{-at}\right)[/tex]

right?

Carry out the derivative and then plug in the information you are given.
 

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