Magnetic dipole in a loop of wire

[Saketh]

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 \mu, 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:
<br /> \varepsilon = -\frac{\partial \Phi_B}{\partial t}<br />

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

Integrating, we get:
<br /> qR = -\Phi_B<br />

Where, I think, \Phi_B 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?

 

[chaoseverlasting]

\phi=B.A
E=\frac{d\phi}{dt}=A\frac{\mu dI}{2adt} as B=\frac{\mu I}{2r}
E=IR=\pi a^2\frac{\mu dI}{2adt}
Solve for I, and magnetic moment is NIA where I A is area vector.

 

[Saketh]

I'm confused.

B=\frac{\mu I}{2r}

How did you get this expression?

 

[Saketh]

By the way, the answer to this problem (from the back of the book) is:

<br /> \mu = \frac{2 a R q}{\mu_0 N}<br />

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