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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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a
2μ0 niπl
b
2μ0 ni2πl
c
2μ0 ni4πl
d
2μ0 niπl

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detailed solution

Correct option is A

Magnetic field due to one side of the square at centre OB1 =μ04π .   2isin 450l⇒ B1 =  μ04π .  2ilHence, magnetic field at centre due to all sidesB=4 B1 =  μ0 2iπlMagnetic field due to n turns,Bnet =nB=μ0 2 niπl = μ0 2 niπl


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