Question

A coil carrying current $l$ has radius r and number of units n. It is rewound so that radius of new coil is $\frac{r}{4}$ and it carries current $l$. 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

The correct option is C $\frac{1}{4}$

Radius and number of turns of the coil initially are $r$ and $n$ respectively.

Thus initial length of the wire, $L=n\left(2\pi r\right)$

Area of the coil, $A=\pi r^2$

Initial magnetic moment, $\mu =nlA=nl\pi r^2$ ..........(1)

Now the same wire is rewound to form the coil of radius $r^{\prime }=\frac{r}{4}$

Let the number of turns in the new coil be $n^{\prime }$

Length of the wire, $L^{\prime }=n^{\prime }\left(2\pi r^{\prime }\right)=\frac{2\pi r}{4}{n}^{\prime }$

But $L=L^{\prime }$, we get, $2\pi rn=\frac{2\pi r}{4}{n}^{\prime }$$⟹n^{\prime }=4n$

Area of the new coil, $A^{\prime }=\pi (r^{\prime }{)}^2=\frac{\pi r^2}{16}$

Magnetic moment of the new coil, ${\mu }^{\prime }=n^{\prime }l{A}^{\prime }=4nl\times \frac{\pi r^2}{16}=\frac{nl\pi r^2}{4}$

$\therefore$ ${\mu }^{\prime }=\frac{\mu }{4}$$⟹\frac{{\mu }^{\prime }}{\mu }=\frac{1}{4}$