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

The magnetic moment \( \mu \) of a current loop is defined as: \[ \mu = I \cdot A \] where \( I \) is the current and \( A \) is the area of the loop. The magnetic field at the centre of the loop due to the current is given by: \[ B = \frac{\mu_0 I A}{2R^2} \] where \( R \) is the radius of the loop. Using the relationship between current and magnetic moment, we can express the current \( I \) in terms of the magnetic moment: \[ I = \frac{\mu}{A} \] Substituting this into the equation for \( B \), we obtain the desired expression for the magnetic moment.