# Magnetic dipole in a loop of wire

• Saketh
In summary, the problem involves a magnetic dipole in a wire loop connected to a galvanometer. The dipole is moved away from the loop and a total charge q passes through the galvanometer. The task is to find the magnetic dipole magnitude \mu in terms of the given variables. The solution involves using Faraday's law and Ohm's law to find the initial magnetic flux and then solving for I using the expression B=\frac{\mu I}{2r}. The final answer is \mu = \frac{2 a R q}{\mu_0 N}.

#### 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.

Ohm's law

## 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?

$$\phi=B.A$$
$$E=\frac{d\phi}{dt}=A\frac{\mu dI}{2adt}$$ 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.

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I'm confused.

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

How did you get this expression?

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.

## 1. What is a magnetic dipole?

A magnetic dipole is a phenomenon in which a magnetic field is generated by a pair of poles with opposite polarities. This is similar to an electric dipole, where a pair of charges with opposite signs create an electric field.

## 2. How is a magnetic dipole created in a loop of wire?

A magnetic dipole can be created in a loop of wire when electric current flows through it. The direction of the current determines the orientation of the magnetic dipole, with the north pole being the end where the current flows in a counterclockwise direction and the south pole being the end where the current flows in a clockwise direction.

## 3. What is the significance of a magnetic dipole in a loop of wire?

The presence of a magnetic dipole in a loop of wire allows for the manipulation of magnetic fields. This is the basis for many applications such as electromagnets, electric motors, and generators.

## 4. How does the strength of a magnetic dipole in a loop of wire change?

The strength of a magnetic dipole in a loop of wire is directly proportional to the amount of current flowing through the wire and the number of turns in the loop. Increasing either of these factors will result in a stronger magnetic dipole.

## 5. Can a magnetic dipole in a loop of wire be turned off?

Yes, a magnetic dipole in a loop of wire can be turned off by stopping the flow of current through the wire. This will cause the magnetic field to dissipate, resulting in the disappearance of the magnetic dipole.