JEE Advanced · Physics · 20. Magnetism & Current
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 \(\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}\)
- C \(-\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}\)
- D \((2 \pi+1) a^{2} I \hat{k}\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{\pi}{2}+1\right) a^{2} I \hat{k}\)
Step-by-step Solution
Detailed explanation
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.
\(\therefore \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}\)

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.
\(\therefore \vec{M}=I a^{2}\left[1+\frac{\pi}{2}\right] \hat{k}\)

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