1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Faraday's law problem

  1. May 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Take a loop of wire with a radius of 2.0 cm. A B field is perpendicular to the area enclosed by the wire. If the field is reduced to zero from 0.55 T in 0.25 s, what is the induced emf between the ends of the wire?

    2. Relevant equations
    B=Magnetic Field
    Flux= BA cos(Theta)
    EMF = Change in Flux/ Change in time

    3. The attempt at a solution

    Trying to find Flux but since its perpendicular to the area, cos90 = 0 and this makes no sense. The answer is suppose to be 0.0028V. Any help would be great, Thanks
  2. jcsd
  3. May 3, 2007 #2
    Your geometry is wrong. You're using the wrong angle.
  4. May 3, 2007 #3
    ok my bad its suppose to be cos0 which is 1
    with that i found Flux to be = (0.55T)(4pi) which is around 6.9T/m^2
    With that 6.9(T/m^2)/0.25(Secs)= 27.6 Volts?
    Like i said answer is suppose to be 0.0027V. Is there something wrong or did i just screw up the units some where?
    Thanks again...
  5. May 3, 2007 #4
    Again the issue is geometry. Look carefully as how you're calculating the area.
  6. May 3, 2007 #5
    The area is just( pi)r^2 right?
  7. May 3, 2007 #6
    Yes, but be careful what you use for "r".
  8. May 3, 2007 #7
    Ohh. I see. its in cm. that should change things abit...Thanks everyone.
  9. May 3, 2007 #8
    Success! I have to ask this question: What about physics did you learn from doing this problem?
  10. May 3, 2007 #9
    ... wat did i learn? I hate physics Tests. =D and UNITS.
  11. May 3, 2007 #10
    Your answer is more or less correct. My point is that there wasn't much physics in this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook