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 \( \mu = I A \), where \( I \) is the current and \( A \) is the loop's area. The magnetic field \( B \) at the center of a circular loop of radius \( r \) due to current \( I \) is \( B = \frac{\mu_0 I}{2r} \). Rearranging for \( I \), we get \( I = \frac{2r B}{\mu_0} \). Substituting this into the magnetic moment equation gives \( \mu = \left( \frac{2r B}{\mu_0} \right) A \). Since \( A = \pi r^2 \), we have \( r = \sqrt{\frac{A}{\pi}} \). Substituting this radius into the equation yields \( \mu = \frac{2B A}{\mu_0} \sqrt{\frac{A}{\pi}} \). Therefore, the magnetic moment of the loop is \( \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).