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A long wire carries a steady curent . It...

A long wire carries a steady curent . 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)(2pil)/(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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