Q.

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

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a

nB

b

n2B

c

2nB

d

2n2B

answer is B.

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Detailed Solution

Let l be the length of the wire. Magnetic field at the centre of the loop isB=μ0I2R∴B=μ0πIl  (∵l=2πR)……….(1) B'=μ0nI2r=μ0nI2l2nπB'=μ0n2πIlFrom eqns. (i) and (ii), we get B'=n2B
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A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be