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_(0)I)/(pi L)`
B. `(2mu_(0)I)/(pi L)`
C. `(sqrt2mu_(0)I)/(pi L)`
D. `(2 sqrt2mu_(0)I)/(pi L)`

Answer

D

Explanation

From figure, `tan 45^(@)= (BE)/(OE) or OE= BE = (L)/(2)`
The magnetic field at point O due to current I through one side AB of the square ABCD will be
`B_(1)=(mu_(0)I)/(4pi ((L)/(2))) [sin 45^(@) + sin 45^(20]= (mu_(0)2I)/(4pi L) [(2)/(sqrt2)]= (mu_(0)2 sqrt2I)/(4pi L) Ox`
The total magnetic field at point O due to current I through all the foure sides of a square will be
`B= 4B_(1)= (mu_(0)2 sqrt2I)/(pi L)`