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A long wire carries a steady current . I...

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

A

`nB`

B

`n^(2)B`

C

`2nB`

D

`2n^(2)B`

Text Solution

Verified by Experts

The correct Answer is:
B

Magnetic field at the centre of single turn loop `B=(mu_(0))/(4pi). (2pii)/r`, magnetic field at the centre of `n`-turn loop `B_(n)=((mu_(0))/(4pi).(2pii)/(r//n))xxn implies B_(n)=n^(2)B`
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