The magnetic field at the point of intersection of diagonals of a square wire loop of side L, carrying a current I is

Options

A. `(mu_0I)/(piL)`
B. `(2mu_0I)/(piL)`
C. `(sqrt2mu_0I)/(piL)`
D. `(2sqrt2mu_0I)/(piL)`

Answer

D

Explanation

`CG=L//2=OG`.

The magnetic field induction at O due to current I through FC of the square FCDE is
`B_1=(mu_0)/(4pi)(I)/(L//2)[sin 45^@+sin 45^@]`
`=(mu_0)/(4pi)(2I)/(L)[1/sqrt2+1/sqrt2]=(mu_02sqrt2I)/(4piL)`
It is acting perpendicular to the plane of paper downwards. Total magnetic field induction at O due to current I through all the four sides of a square will be
`B=4B_1=4xx(mu_02sqrt2I)/(4piL)=(mu_02sqrt2I)/(piL)`