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 = \( B \)
• Objective: To demonstrate that \( M = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \)

Derivation Steps:

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

For a circular loop of radius \( r \) with current \( I \), the magnetic field at its centre is given by: \[ B = \frac{\mu_0 I}{2r} \]

2. Area of the Loop:

The area \( A \) of a circle with radius \( r \) is defined as: \[ A = \pi r^2 \]. Therefore, the radius can be expressed as: \[ r = \sqrt{\frac{A}{\pi}} \]

3. Isolating Current in the Magnetic Field Equation:

Rearranging the magnetic field formula \( B = \frac{\mu_0 I}{2r} \) to solve for the current \( I \): \[ I = \frac{2Br}{\mu_0} \]

4. Calculating the Magnetic Moment of the Loop:

The magnetic moment \( M \) of a current loop is calculated as the product of the current and the area: \[ M = I \cdot A \] Substitute the expression for \( I \) derived previously: \[ M = \left( \frac{2Br}{\mu_0} \right) \cdot A \] Now, substitute the expression for \( r \) in terms of \( A \): \[ M = \frac{2B}{\mu_0} \cdot A \cdot \sqrt{\frac{A}{\pi}} = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]

✔ Result:

The magnetic moment of the circular loop has been shown to be: \[ M = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]