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Homework Help: The electric generator

  1. Jul 29, 2005 #1
    The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 5.4 m. The magnetic field of the generator is 0.10 T, and the coil rotates at an angular speed of 35 rad/s. What is the peak emf of this generator?

    so i have:
    r= .14m
    L= 5.4m
    B= .10T
    w=35 rad/s

    now i thought i would do:
    emf= BLv
    v=rw... v= .14m(35 rad/s)
    emf= (.10T)(5.4m)(.049m/s)= .02646
    and peak emf= (square root 2)(emf)= .0374...

    this problem is wrong the way i tried it, but i'm not sure what i should do differently
  2. jcsd
  3. Jul 30, 2005 #2


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    Homework Helper
    Gold Member

    The induced EMF (across the ends of the rod) due to the motion of a rod of length 'l' and velocity 'v', in the presence of a magnetic field of strength 'B' is Blv. So this formula is not applicable here as there is a rotating coil and not a rod.

    To solve this problem, go from the definition of Farady's law.
    By Farady's law, Emf induced = -d(Magnetic Flux)/dt

    Let the magnetic field make an angle theta with the area vector of the loop at any time 't' such that at t=0, theta=0.
    So the Magnetic flux enclosed by the loop is = [itex] n B.A [/itex]
    where n is the number of loops, B is the magnetic field and A is the area of the loop.

    [tex] = (n)(B)(A)(\cos\theta) [/tex]

    So, the EMF induced will be

    [tex] =\frac {-d[(n)(B)(A)(\cos\theta)]}{dt} [/tex]

    From this, can you calculate the EMF as a function of time and from that the peak value?
    (You will have to find the relation between 'theta' and 't' as well as the value of n)
    Last edited: Jul 30, 2005
  4. Jul 30, 2005 #3
    The coil is rotating in the field. The flux is thus changing and this causes the electric field in the coil.

    [tex] \Phi = AB [/tex], B is constant but A is changing. Can you find A as a function of time?

    [tex] E = -N \frac{d\Phi}{dt} [/tex], so you will also need to find N - the number of layers in the coil.

    Just find [tex] \frac{dA}{dt}[/tex] and the biggest problem is probably solved.
  5. Jul 30, 2005 #4
    thanks guys... you really helped me
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