A coil is carrying a current of 10 A and has radius 10 cm and number of turns 500. It is rewound to make a new coil of radius 5 cm and it carries same current 10 A. The ratio of magnetic moment of original coil to that of new coil is

Correct option is D

​$$\begin{aligned}&\textbf{Given:} \\&\text{Original coil:} \\&\quad \text{Number of turns } N_1 = 500 \\&\quad \text{Radius } R_1 = 10\, \text{cm} = 0.1\, \text{m} \\&\quad \text{Current } I = 10\, \text{A} \\\\&\text{Magnetic moment of original coil} \\&\quad M_1 = N_1 \cdot I \cdot \pi R_1^2 = 500 \cdot 10 \cdot \pi \cdot (0.1)^2 = 5 \times 10^5 \pi \\\\&\text{New coil:} \\&\quad \text{Radius } R_2 = 5\, \text{cm} = 0.05\, \text{m} \\&\quad \text{Same length of wire } \Rightarrow N_1 \cdot 2\pi R_1 = N_2 \cdot 2\pi R_2 \\\\&\text{So,} \\&\quad 500 \cdot 0.1 = N_2 \cdot 0.05 \Rightarrow N_2 = 1000 \\\\&\text{Magnetic moment of new coil} \\&\quad M_2 = N_2 \cdot I \cdot \pi R_2^2 = 1000 \cdot 10 \cdot \pi \cdot (0.05)^2 = 25 \times 10^4 \pi \\\\&\text{Ratio of magnetic moments} \\&\quad \frac{M_1}{M_2} = \frac{5 \times 10^5 \pi}{2.5 \times 10^5 \pi} = \frac{2}{1}\end{aligned}$$​​