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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:
    [tex]
    \varepsilon = -\frac{\partial \Phi_B}{\partial t}
    [/tex]

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

    Integrating, we get:
    [tex]
    qR = -\Phi_B
    [/tex]

    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]\phi=B.A[/tex]
    [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:

    [tex]
    \mu = \frac{2 a R q}{\mu_0 N}
    [/tex]

    I'm still confused -- chaoseverlasting, if I do what you said, I'm getting an exponential growth function, which doesn't make sense.
     
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