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Homework Help: Faraday's Law of Electromagnetic Induction

  1. Feb 20, 2005 #1
    A constant magnetic field passes through a single rectangular loop whose dimensions are 0.348 m x 0.593 m. The magnetic field has a magnitude of 2.26 T and is inclined at an angle of 66.0° with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of 0.475 s, what is the magnitude of the average emf induced in the loop?
    (b) If the magnetic field remains constant at its initial value of 2.26 T, what is the magnitude of the rate at which the area should change so that the average emf has the same magnitude?

    I know how to figure out part a, its part b that confused me.
    a) E=NAcos(theta)((B1-B2)/(t1-t2))
    =3.99*10^-1 V

    b) E=N(change in flux/change in time) and (change in flux)=BAcos(theta)
    E=N(BAcos(theta)/(change in time))
    E=NA(cos(theta))(change in B/change in time)

    I think I may be going about an equation wrong, and I cant distinguish when to focus on a change in B and change in area. because I worked with B in the previous question. I dunno, I'm just confused.
  2. jcsd
  3. Feb 20, 2005 #2


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    [tex] e=-\frac{d\Phi}{dt} [/tex]...If you have found the correct "e" at point a),the for point b),u need to compute
    [tex] \frac{dA}{dt} [/tex]...

  4. Feb 20, 2005 #3
    (change in A/change in time) = ((m)(m)) / (NBcos(theta))

    It's not clear to me what I did wrong with this.
    okay, nevermind.
    V=B(change in area/change in time)
    is there some special thing im overlooking cuz plug and
    chugging into equations I know arent helping me understand
    Last edited: Feb 21, 2005
  5. Feb 21, 2005 #4


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    [itex] d\Phi =\vec{B}\cdot\vec{n} dA=B\cos 66deg \ dA [/tex]

    Divide through "-dt"...
    [tex] e=-B\cos 66deg \ \frac{dA}{dt} [/tex]

    And now express [itex] \vec{dA}{dt} [/itex] as a function of B,e,cos 66°...

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