The magnetic field at the point O in the following current carrying square loop at the centre will be

Options

A. `12xx 10^(-5)` T
B. `4xx 10^(-5)` T
C. `4 sqrt(2) xx 10^(-5)` T
D. `2 sqrt(2) xx 10^(-5)` T

Answer

C

Explanation

Given, current through the square loop is 2A.

The given situation is shown in the figure. Magnetic field due to wire AB at the centre O is
`B_(AB) = (mu_(0))/(4pi). (l)/( r) [ sin 45^(@) + sin 45^(@)]`
=`10^(- 7) xx (2)/(2xx10^(-2)) [ (1)/(sqrt(2))+(1)/(sqrt(2))]`
=` sqrt(2) xx 10^(-5) `T [Downward to the plane of square loop]
Since, magnetic field due to side AB,BC,CD and DA is same in magnitude and direction. Hence, net magnetic field at O
`B =4 xx B_(AB) = 4 sqrt(2) xx 10^(-5) T`
Wheatstone bridge in balanced condition maintains `(P)/(Q)=(R )/(S)`
i.e. no current flows through the galvanometer (i.e. `i_(g) =0)`