# Magnetic dipole in a loop of wire

1. Mar 13, 2007

### Saketh

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 $\mu$, the magnetic dipole magnitude, in terms of the given variables.

2. Relevant equations

Ohm's law

3. The attempt at a solution

$$\varepsilon = -\frac{\partial \Phi_B}{\partial t}$$

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, $\Phi_B$ 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?

2. Mar 14, 2007

### 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.

Last edited: Mar 14, 2007
3. Mar 14, 2007

### Saketh

I'm confused.

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

How did you get this expression?

4. Mar 14, 2007

### Saketh

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.