Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Faraday's law of electomagnetic induction

  1. Jul 29, 2005 #1
    A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.60 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 x 10-2 ohm/m. What is the average electrical energy dissipated in the resistance of the wire.

    ok so this is what i have:
    r= 13cm= .13m
    change in t= .45s
    B= .60T
    restitance per unit length= 3.3 x 10^-2
    N= 1

    then i tried:
    magnetic flux= BAcos= (.60T)(530.9m^2)= 318.6
    emf= -N(change in magnetic flux/ change in time)= -1(318.6/.45s)= -708

    here is where i get really confused:
    R= restitance per unit length (L/A)

    does this mean i should do:
    R= 3.3 x 10-2 ohm/m (2 pi .13)/(.13^2 pi)???

    or is it just R= 3.3 x 10-2 ohm/m (.13m)??? because the units work out this way...

    once i figure that out i would do:
    I=emf/R ---->
    P=I(emf) ------->
    E=Pt and that would give me energy
  2. jcsd
  3. Jul 30, 2005 #2


    User Avatar
    Homework Helper

    you are guven resistance per unit length
    hence 2Pi r length will have resistance 3.3x10^-2*(2Pi -0.13)
    Last edited: Jul 30, 2005
  4. Jul 30, 2005 #3
    won't that give me units of ohm/m^2 ? i thought i was supposed to be getting rid of the m's because the R is in units of ohm...
    i'm still very confused about this problem.
  5. Jul 30, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    [tex] R= 3.3 \times 10^{-2} \left(\frac{ohm}{m}\right) \cdot \left(2 \pi \times 0.13m \right) [/tex]

    This has units of ohms. The last term is the circumference of the circle, i.e., the length of the wire.
  6. Jul 30, 2005 #5
    ok i think i finally understand how to find R now...
    R= .027

    once i figured that out, i did:
    I=emf/R = (-708)/.027 = -26266.1
    P=I(emf) = (-26266.1)(-708) = 18596420.37W
    E=Pt = (18596420.37W)(.45s) = 8368389.167J and that should give me the electrical energy dissipated in the wire right???

    but this problem still isn't working out and i'm not sure where i went wrong
  7. Jul 30, 2005 #6


    User Avatar

    Check your formula for the magnetic flux. Is the magnetic field always at 0.6 Tesla? Also, pay attention to the fact that the question is asking for the average energy dissipated.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook