If a straight conductor of length `40cm` bent in the form of a square and the current `2A` is allowed to pass through square, then find the magnetic induction at the centre of the square loop.

Explanation

`B_(n et)=4B_(si de)`
`B_(n et)=4(mu_(0))/(4pi)xx(I)/(L//2)(sin45^(@)+sin45^(@))`
`=4X(mu_(0))/(4pi)xx(I)/(L//2)(sqrt(2))=(mu_(0))/(4pi)(8sqrt(2I))/(L)`
`=10^(-7)xx8sqrt(2)xx2xx10=16sqrt(2)muT`