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
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Answer 1

The magnetic field will be equal to the option a).

First we will find the magnetic field due to one side of the square at the center.

B1 = μ/4π × 2isin45°/(a/2)

where a is the side , i is the current flowing .

B1= μ/4π × 2√2i/a

Now , the magnetic field at the center due to all sides -

B = 4B1

=> μ/π × 2√2 i/a (a=2l)

=>μ/π × √2 i/l

For n turns the net magnetic field ,

Bnet =nB

=>√2μ/π × n i/l