Question

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

पर्याय

• `(sqrt2 mu_0 "nI")/(pil)`

• `(sqrt2 mu_0 "nI")/(2pil)`

• `(sqrt2 mu_0 "nI")/(4pil)`

• `(2 mu_0 "nI")/(pil)`

Answer 1

The magnetic field at the centre of coil of n turns, bent in the form of a square of side `2l`, carrying current i, is `(sqrt2 mu_0 "nI")/(pil)`.

Explanation:

Using Biot Savart's law,

Magnetic field due to one side of the square at centre O

`"B"_1 = (mu_0"i")/(4 pil) xx ("sin"45°+ "sin"45°)`

`= mu_0/(4pi) xx (sqrt2 "i")/l`

Hence magnetic field at centre due to all sides,

`"B" = 4"B"_1 = (mu_0 (sqrt2) "i")/(pil)`

Magnetic field due to n turns

`"B"_"net" = "nB" = (sqrt2 mu_0 "nI")/(pil)`