Question

A coil carrying current ‘$I$’ has radius ‘$r$’ and number of turns ‘$n$’. It is rewound so that radius of new coil is $\frac{'r'}{4}$ 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,
$M_{0}=n_{1} I A=n I\left(\pi r^{2}\right)\,\,\,\,\,...(i)$
Magnetic moment of the new coil,
$M_{n}=n_{2} I A=n_{2} I\left(\pi \frac{r^{2}}{16}\right) \,\,\,\,\,...(ii)$
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=4 n \,\,\,\,\,\,...(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}=4 n\right] $
$\Rightarrow \frac{M_{n}}{M_{0}}=\frac{1}{4}$