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

Options

A. `a^(2) I hat k`
B. `(pi/2 + 1) a^(2) I hat k`
C. `-(pi/2 + 1) a^(2) I hat k`
D. `(2pi + 1) a^(2) I hat k`

Answer

b

Explanation

If we curl our right hand fingers in the sense of current then thumb points outward. Hence direction of magnetic moment is outward, which is positive Z-axis.
The magnitude of magnetic moment is equal to current multiplied with area. The area of given coil is equal to the area of square of side , and four semicircles of radius a/2.
`Lambda = a^(2) | 4 xx (0.5) pi (a//2)^(2) = a^(2) + pi a^(2) //2 - (pi //2 + 1) a^(2)`
Magnetic Moment, `vec M = I A hat k`
Hence, the magnetic moment can be written as
`(pi/2 + 1) a^(2) I hat k`
Thus, option (b) is correct.