An infinitesimal bar magnet of dipole moment `M` is pointing and moving with speed `v` in the `x`-direction. A closed circular conducting loop of radius a and negligible self-inductance lies in the `y-z` plane with its centre at `x=0` and its axis coinciding with `x`-axis. find the force opposing the motion of the magnet, if the resistance of the loop is `R` . Assume that the distance `x` of the magnet from the centre of the loop is much greater than a .

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Magnetic Field Due to the Bar Magnet
The magnetic field \( B \) at a distance \( x \) from the bar magnet along its axis is given by the formula:

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{x^3} \]

Where:
- \( \mu_0 \) is the permeability of free space,
- \( M \) is the magnetic dipole moment of the bar magnet,
- \( x \) is the distance from the magnet to the center of the loop.

### Step 2: Calculate the Induced EMF in the Loop
As the magnet moves with speed \( v \), the magnetic field at the loop changes with time. The change in magnetic flux \( \Phi \) through the loop can be expressed as:

\[ \Phi = B \cdot A \]

Where \( A \) is the area of the loop, given by \( A = \pi a^2 \). The induced EMF \( \mathcal{E} \) in the loop can be calculated using Faraday's law of electromagnetic induction:

\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

Using the chain rule, we can express the time derivative of the magnetic flux as:

\[ \mathcal{E} = -\frac{dB}{dx} \cdot \frac{dx}{dt} = -\frac{dB}{dx} \cdot v \]

### Step 3: Calculate the Derivative of the Magnetic Field
Now, we need to find \( \frac{dB}{dx} \):

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{x^3} \]

Taking the derivative with respect to \( x \):

\[ \frac{dB}{dx} = -\frac{3\mu_0}{4\pi} \cdot \frac{2M}{x^4} \]

### Step 4: Substitute into the EMF Equation
Substituting \( \frac{dB}{dx} \) into the EMF equation gives:

\[ \mathcal{E} = -\left(-\frac{3\mu_0}{4\pi} \cdot \frac{2M}{x^4}\right) \cdot v = \frac{3\mu_0 M v}{2\pi x^4} \]

### Step 5: Calculate the Induced Current
Using Ohm's law, the induced current \( I \) in the loop can be calculated as:

\[ I = \frac{\mathcal{E}}{R} = \frac{3\mu_0 M v}{2\pi R x^4} \]

### Step 6: Calculate the Force Opposing the Motion
The force \( F \) opposing the motion of the magnet can be calculated using the formula:

\[ F = I \cdot B \cdot 2\pi a \]

Substituting the values of \( I \) and \( B \):

\[ F = \left(\frac{3\mu_0 M v}{2\pi R x^4}\right) \cdot \left(\frac{\mu_0}{4\pi} \cdot \frac{2M}{x^3}\right) \cdot 2\pi a \]

This simplifies to:

\[ F = \frac{3\mu_0^2 M^2 v a}{4 R x^7} \]

### Final Result
Thus, the force opposing the motion of the magnet is:

\[ F = \frac{3\mu_0^2 M^2 v a}{4 R x^7} \]