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

  $a^{2} I \hat{k}$
• Correct answer
  B

  $\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$
• C

  $-\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}$
• D

  $(2 \pi+1) a^{2} I \hat{k}$

Solution

Magnetic moment of a current carrying loop $\vec{M}=N I \vec{A}$ Here $N=1, A=a^{2}+2 \pi\left(\frac{a}{2}\right)^{2}=a^{2}\left[1+\frac{\pi}{2}\right]$ From Screw law, direction of $m$ is outward or in +ve z-direction. $\therefore \quad \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}$

Hence, the correct answer is option: 'B'.