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
n B
B
`n^2B`
C
`2 n B`
D
`2 n^2 B`
Text Solution
Verified by Experts
The correct Answer is:
B
For same length , `B prop n^2` `therefore (B_2)/(B_1)=((n_2)/(n_1))^2` `B_2=((n)/(1))^2xxB=n^2B`.
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