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: \[ \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]

Hint:

The magnetic moment of a current loop is the product of the current and the area of the loop. The magnetic field produced at the center of the loop depends on both the current and the area.

Answer 1

The magnetic moment \( \mu \) of a current loop is defined as: \[ \mu = I A \] where \( I \) is the current and \( A \) is the area of the loop. The magnetic field \( B \) produced by a current \( I \) at the centre of a circular loop of radius \( r \) is given by: \[ B = \frac{\mu_0 I}{2r} \] From this, we can solve for \( I \): \[ I = \frac{2r B}{\mu_0} \] Now, substitute this value of \( I \) into the equation for the magnetic moment: \[ \mu = \left( \frac{2r B}{\mu_0} \right) A \] Since the area \( A \) of the loop is related to the radius by \( A = \pi r^2 \), substitute \( r = \sqrt{\frac{A}{\pi}} \) into the equation: \[ \mu = \frac{2B A}{\mu_0} \sqrt{\frac{A}{\pi}} \] Thus, the magnetic moment of the loop is: \[ \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \]