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: Magnetic dipole in a loop of wire

  1. Mar 13, 2007 #1
    This isn't a homework problem, I'm just doing this as practice.

    1. The problem statement, all variables and given/known data

    A magnetic dipole is oriented in a loop of wire of N turns and radius a so that the dipole vector is parallel to the normal of the loop. The loop is connected to a galvanometer, and the active resistance of the circuit is R.

    The dipole is moved away from the loop, and in the process a total charge q passes through the galvanometer. Find [itex]\mu[/itex], the magnetic dipole magnitude, in terms of the given variables.

    2. Relevant equations

    Ohm's law
    Faraday's law of induction

    3. The attempt at a solution

    From Faraday's law:
    \varepsilon = -\frac{\partial \Phi_B}{\partial t}

    From Ohm's law:
    \varepsilon = IR = \frac{d q}{dt}R = -\frac{\partial \Phi_B}{\partial t}

    Integrating, we get:
    qR = -\Phi_B

    Where, I think, [itex]\Phi_B[/itex] represents the initial magnetic flux.

    Now I have two questions:
    1. Is what I have done so far correct?
    2. How am I supposed to find the magnetic flux?
  2. jcsd
  3. Mar 14, 2007 #2
    [tex]E=\frac{d\phi}{dt}=A\frac{\mu dI}{2adt}[/tex] as [tex]B=\frac{\mu I}{2r}[/tex]
    [tex]E=IR=\pi a^2\frac{\mu dI}{2adt}[/tex]
    Solve for I, and magnetic moment is NIA where I A is area vector.
    Last edited: Mar 14, 2007
  4. Mar 14, 2007 #3
    I'm confused.

    [tex]B=\frac{\mu I}{2r}[/tex]

    How did you get this expression?
  5. Mar 14, 2007 #4
    By the way, the answer to this problem (from the back of the book) is:

    \mu = \frac{2 a R q}{\mu_0 N}

    I'm still confused -- chaoseverlasting, if I do what you said, I'm getting an exponential growth function, which doesn't make sense.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook