1. Not finding help here? Sign up for a free 30min 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!

Induced current

  1. May 15, 2006 #1
    Here is a question I have on a practice exam that I am having a littel trouble with.
    A plane loop of wire consisting of a single turn if radius 5cm is perpendicular to a magnetic field that increases uniformly in magnitude from 0.5T to 2.5T in 1.5 seconds what is the resluting indiced current if the coil has a resistance of 2ohms?

    I am not sure where to procede I need something that leads me to current so finding a voltage would be nice because I already have the resistance.
    perhaps using faradays law?
     
  2. jcsd
  3. May 15, 2006 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Yes. Can you post your attempt?
     
  4. May 15, 2006 #3
    Just use
    [tex]\epsilon = \frac{d\phi}{dt}[/tex]
     
  5. May 15, 2006 #4

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Come on arun, let the original poster show some work!

    There's no fun in giving away the answer just like that, is there? The original poster will learn more if he/she figures it out for themselves don't you think :smile:?
     
  6. May 15, 2006 #5
    Well you already said that Faraday's laws are the way to go. I just wrote the mathematical form.
    Besides I don't think there is much to do in this problem, otherwise I usually look at the working.

    Cheers,
    Arun
     
  7. May 17, 2006 #6
    OK I am back. It turns out this problem was on the test with one variation the radius went to 15cm. I used faradays law but got the answer wrong here is what I did.

    emf=(pie(.15^2)/1.5=.0471 volts

    and then to find the current I=V/R=.0471/2=.0236
     
  8. May 17, 2006 #7

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The flux is the area times the B field (times a cosine but here the cos gives 1) . So [itex] {d \phi \over dt} = A {\Delta B \over \Delta t} [/itex]. Looks like you forgot the change of B field, [itex] \Delta B [/itex]

    Patrick
     
  9. May 17, 2006 #8
    Crap, so the change of B would be two so if I put
    emf=(2pie(.15^2)/1.5=.942 volts
    and then current would be .942/2=.0471 ohms

    Does the answer look good now?
     
  10. May 17, 2006 #9

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I get 10 times smaller than your results ( I used 2 T* pi (0.15 m)^2/1.5 s also and got 0.0942 T m^2/s)
     
  11. May 18, 2006 #10
    I get the same thing as you...apparently I wasnt paying atttention before.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Induced current
  1. Induced Current (Replies: 1)

  2. Induced current (Replies: 7)

  3. Induced current (Replies: 3)

  4. Induced Current (Replies: 3)

  5. Induced Current (Replies: 5)

Loading...