Question

A coil carrying current ‘$I$’ has radius ‘$r$’ and number of turns ‘$n$’. It is rewound so that radius of new coil is $\frac{{}^{\prime }{r}^{\prime }}{4}$ and it carries current ‘$I$’. The ratio of magnetic moment of new coil to that of original coil is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

Magnetic moment of the original coil,
$M_0=n_{1}IA=nI\left(\pi r^2\right)...(i)$
Magnetic moment of the new coil,
$M_n=n_{2}IA=n_{2}I\left(\pi \frac{r^2}{16}\right)...(ii)$
Equating length of the wires we have,
$n_1\times 2\pi r_1=n_2\times 2\pi r_2$
$\Rightarrow n\times 2\pi r=n_2\times 2\pi \times \frac{r}{4}$
$\Rightarrow n_2=4n...(iii)$
Now from Eqs. (i) and (ii), we get
$\frac{M_0}{M_n}=\frac{n}{n_2}\times 16=\frac{1}{4}\times 16=4\left[\because n_2=4n\right]$
$\Rightarrow \frac{M_n}{M_0}=\frac{1}{4}$