Question

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.

Answer 1

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)))`