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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:
A
2nB
B
$2n^2B$
C
nB
D
$n^2B$
Explanation
n turns and radius R/n give $n^2$ times the field.
Detailed Solution
Since $l = 2\pi R = n(2\pi r) \Rightarrow r = \frac{R}{n}$
For one turn: $B = \frac{\mu_0i}{2R}$
For n turns: $B' = \frac{\mu_0ni}{2r} = \frac{\mu_0n^2i}{2R} = n^2B$
For one turn: $B = \frac{\mu_0i}{2R}$
For n turns: $B' = \frac{\mu_0ni}{2r} = \frac{\mu_0n^2i}{2R} = n^2B$