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. $(\frac{\pi }{2}+1)a^{2}I\hat{k}$
• C. $-(\frac{\pi }{2}+1)a^{2}I\hat{k}$
• 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)
$=2\pi (\frac{a}{2}{)}^2+a^2=(\frac{\pi }{2}+1)a^2$
$|M|=IA=(\frac{\pi }{2}+1)a^{2}I$
From screw law direction of M is outwards or in positive z-direction.
$\therefore M=(\frac{\pi }{2}+1)a^{2}Ik$