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

• A. $\frac{N}{8}$
• B. $\frac{N}{4}$
• C. 2N
• D. 4 N

Answer 1

The correct option is A $\frac{N}{8}$

$\frac{B_c}{M}=\frac{\frac{{\mu }_{0}ni}{2R}}{\pi R^{2}ni}=\frac{{\mu }_0}{2\pi R^3}=NN\alpha \frac{1}{R^3};N^1\alpha \frac{1}{(2R{)}^3}{N}^1\alpha \frac{1}{8R^3}\frac{N^1}{N}=\frac{\frac{1}{8R^3}}{(\frac{1}{R^3}}=\frac{1}{8};N^1=\frac{N}{8}$