A non - popular loop of conducting wire carrying a current `I` is placed as shown in the figure . Each of the straight sections of the loop is of the length ` 2a`. The magnetic field due to this loop at the point `P(a, 0 ,a)` points in the direction

Options

A. `1/(sqrt(2))(-hatj+hatk)`
B. `1/(sqrt(3))(-hatj+hatk+hati)`
C. `1/(sqrt(3))(hati+hatj+hatk)`
D. `1/(sqrt(2))(hati+hatk)`

Answer

D

Explanation

The magnetic field at `P(a,0,a)` due to the loop is equal to the vector sum of the magnetic fields produces by loops `ABCDA` and `AFEBA` as shown in the figure. Magnetic field due to loop `ABCDA` will be along `hati` and due to loop `AFEBA`, along `hatk`. Magnitude of magnetic field due to both the loops will be equal. Therefore, direction of resultant magnetic field will be `1/(sqrt(2))(hati+hatk)`.