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Induction Trouble

  1. Mar 10, 2004 #1
    I think I might be confused on some conceptual issues on this problem as I can't really come up with a way to get a numerical answer on anything in this problem. Please help me with any guidance or clarification.

    The Problem:

    A square (2.3 cm on each side) wire loop lies 9.0 cm away from a long, straight wire. The square lies in the same plane as the long wire. The resistance of the wire in the loop is 79 ohms. The long wire carries a current of 6.8 A.
    a.) What emf is induced in the loop when it's at rest?
    b.) What direction is the current induced in the loop (also when at rest)?
    c.) If the loop begins moving away from the wire, in what direction is the current induced in this loop?

    My Reasoning/Attempt:

    Well first I found the magnetic field created by the wire using the equation B = µ*I/(2*pi*0.09 m) and found that the magnetic field was 1.51 x 10^-5 T.
    This is where I got stumped because I started thinking about the problem a little more.
    For a.) I figured if there is no velocity since the loop is at rest even though there is a magnetic field there should be no emf. So am I right to say there is 0 V for the emf at rest?
    For b.) I said there should be no current because there is no emf at rest so the current is 0 amps.
    For c.) I said the current of the loop will be induced in the opposite direction as the current in the wire because current in wires of the opposite direction repel each other.

    Does this reasoning look right or is there actually a numerical representation required for this question?
     
  2. jcsd
  3. Mar 10, 2004 #2
    I'd say you're right. Since magnetic flux through the loop is a constant, no emf will be induced. As for the third question: Lanz' law says that the induced emf will be such that it opposes the change that caused it. After twisting my rigth for a while I, too, come to the conclusion that induced current will flow in the opposite direction..
     
  4. Mar 10, 2004 #3
    Thank you for your help :)
     
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