Question

A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

Hint:

The magnetic moment of a current loop is directly proportional to both the current and the area of the loop.

Answer 1

Magnetic Moment of a Circular Current-Carrying Loop 

Given:

• Area of the loop = \( A \)
• Magnetic field at the centre = \( B \)
• We are to show: \( M = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \)

Step-by-Step Derivation:

1. Magnetic Field at the Centre of a Circular Loop:

The magnetic field at the centre of a circular loop of radius \( r \) carrying current \( I \) is: \[ B = \frac{\mu_0 I}{2r} \]

2. Area of the Loop:

The area \( A \) of a circle of radius \( r \) is: \[ A = \pi r^2 \Rightarrow r = \sqrt{\frac{A}{\pi}} \]

3. Rearranging Magnetic Field Equation:

From \( B = \frac{\mu_0 I}{2r} \), solve for \( I \): \[ I = \frac{2Br}{\mu_0} \]

4. Magnetic Moment of the Loop:

Magnetic moment \( M \) of a current loop is given by: \[ M = I \cdot A \] Substituting \( I \) from above: \[ M = \left( \frac{2Br}{\mu_0} \right) \cdot A \] Now substitute \( r = \sqrt{\frac{A}{\pi}} \): \[ M = \frac{2B}{\mu_0} \cdot A \cdot \sqrt{\frac{A}{\pi}} = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]

✔ Conclusion:

The magnetic moment of the circular loop is: \[ M = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \] as required.