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Magnetic dipole in a loop of wire

  • Thread starter Saketh
  • Start date
  • #1
261
2
This isn't a homework problem, I'm just doing this as practice.

Homework Statement



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.

Homework Equations



Ohm's law
Faraday's law of induction

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?
 

Answers and Replies

  • #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:
  • #3
261
2
I'm confused.

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

How did you get this expression?
 
  • #4
261
2
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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