Question

A loop carrying current $I$ lies in the $x-y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is

• A. $a^2 I \hat{k}$
• B. $\bigg( \frac{\pi }{2}+1 \bigg) a^2 I \hat{k}$
• C. $ -\bigg( \frac{\pi }{2}+1 \bigg) a^2 I \hat{k}$
• D. $ (2 \pi +1)a^2 I \hat{k}$

Answer 1

Answer: (B)

Area of the given loop is
A = (area of two circles of radius $ \frac{a}{2}$ and area of a square of side a)
$ \, \, \, \, \, \, \, \, \, \, = 2 \pi \bigg( \frac{a}{2} \bigg) ^2+a^2 = \bigg( \frac{\pi}{2}+1 \bigg) a^2$
$ \, \, \, \, \, \, \, \, | M| = IA = \bigg( \frac{\pi}{2}+1 \bigg) a^2 I $
From screw law direction of M is outwards or in positive z-direction.
$\therefore \, \, \, \, \, \, \, \, M = \bigg( \frac{\pi}{2}+1\bigg) a^2I k $