Magnetic field at the centre of coil of `n` turns, bent in the form of a square of side `2l`, carrying current `i`, is

Options

A. `(sqrt2 mu_(0) nI)/(pi l)`
B. `(sqrt2 mu_(0) nI)/(2pi l)`
C. `(sqrt2 mu_(0) nI)/(4pi l)`
D. `(2mu_(0) n I)/(pi l)`

Answer

A

Explanation

Magnetic field due to finite wire - `B = (mu_(0) i)/(4pi a) (sin theta_(1) + sin theta_(2))`
B due to one side `B_(1) = (mu_(0) i)/(4pi a) (sin theta _(1) + theta_(2)) = (mu_(0) i)/(4pi l) [ (1)/(sqrt2) + (1)/(sqrt2)] = (mu_(0)isqrt2)/(4pi l)`

B due to square `" " [a = sqrt((sqrt2l)^(2) - l^(2)) = l]`
`B = (mu_(0) i "n" 4 sqrt2)/(4pi l) " "[ n to ` No. of turns]
`implies B = (mu_(0) i "n" sqrt2)/(pi l)`