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

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

Initially, the radius of the coil $r_1=r$

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

Thus initial magnetic moment of the coil ${\mu }_1=nIA=2\pi nI{r}^2$

Now the coil is rewound to form a new coil of radius $r=\frac{r}{4}$ having $n^{\prime }$ turns.

Length of the coil before and after its is rewounded remains the same.

$\therefore$ $n\left(2\pi r\right)=n^{\prime }(2\pi \times \frac{r}{4})$

$⟹$ $n^{\prime }=4n$

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

Magnetic moment of new coil ${\mu }^{\prime }=n^{\prime }I{A}^{\prime }$

$\therefore$ ${\mu }^{\prime }=\left(4n\right)I\times \frac{\pi r^2}{16}=\frac{\pi nI{r}^2}{4}$

Thus ratio of magnetic moments $\frac{{\mu }^{\prime }}{\mu }=\frac{\pi n{r}^2/4}{\pi n{r}^2}$

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