A current loop is bent in the form of a rectangule of sides a and b as shown. The current flowing through the loop is i and its direction is shown in the figure. Calculate the magnetic field at the centre of the rectangle

Strategy : Due to each side of the rectangle, field at O is perpendicular to the plane of paper adn outwards (you can check it using Right Hand Thumb Rule). So the fields can be added up to find net magnetic field.

Explanation

Due to straight conductor AD, magnetic field is given by
`B_(AD) = (mu_(0)i)/(4pi(b//2)) [sin theta + sin theta]`
[Using `B = (mu_(0)i)/(4pi r) (sin phi_(1) + sin phi_(2))`]
As `sin theta = (AD)/(AC) = (a)/(sqrt(a^(2) + b^(2)))`
`:. B_(AD) = (mu_(0)i)/(pi b) ((a)/(sqrt(a^(2) + b^(2))))`
SImilarly, `B_(AB) = (mu_(0)i)/(pi a) ((b)/(sqrt(a^(2) + b^(2))))` (Hint : Interchange a and b)
`B_(BC) = (mu_(0)i)/(pi b) ((a)/(sqrt(a^(2) + b^(2))))` (same as `B_(AD)`)
and `B_(CD) = (mu_(0)i)/(pi a) ((b)/(sqrt(a^(2) + b^(2))))` (same as `B_(AB)`)
Net magnetic field `B = B_(AB) + B_(CD) + B_(DA) = (2mu_(0)i)/(pi) [(a)/(bsqrt(a^(2) + b^(2))) + (b)/(a sqrt(a^(2) + b^(2)))]`
`= (2mu_(0)i)/(pi ab) (sqrt(a^(2) + b^(2)))`