A wire in the form of a circular loop of one turn carrying a current produces a magnetic field B at the centre. If the same wire is looped into a coil of two turns and carries the same current, the new value of magnetic induction at the centre is

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answer is D.

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Detailed Solution

$B = \frac{μ_{0} i}{2 r} = \frac{μ_{0} i}{2 (\frac{l}{2 π)}} = \frac{{πμ}_{0} i}{l} N o w 2 πr' n = l \Rightarrow r' = \frac{l}{2 πn} Therefore, B' = \frac{μ_{0} ni}{2 r'} = \frac{μ_{0} πi}{l} n^{2} = {Bn}^{2} = B {(2)}^{2} = 4 B$

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