The ratio of magnitude of magnetic field at the centre of a circular current car

Question

The ratio of magnitude of magnetic field at the centre of a circular current carrying coil to its magnetic moment is N. If the current and radius both were doubled the ratio will become

Options:

A . N8

B . N4

C . 2N

D . 4 N

Answer: Option A
:
A

\frac{B_{c}}{M} = \frac{\frac{μ_{0} n i}{2 R}}{π R^{2} n i} = \frac{μ_{0}}{2 π R^{3}} = N N α \frac{1}{R^{3}}; N^{1} α \frac{1}{(2 R)^{3}} N^{1} α \frac{1}{8 R^{3}} \frac{N^{1}}{N} = \frac{\frac{1}{8 R^{3}}}{(\frac{1}{R^{3}}} = \frac{1}{8}; N^{1} = \frac{N}{8}

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