Question

The field normal to the plane of wire of $n$ turns and radius $r$ which carries a current $i$ is measured on the axis at a small distance $h$ from the centre of the coil. This is smaller than the field at the centre by the fraction

• A. $\frac{2}{3}\frac{h^2}{r^2}$
• B. $\frac{3}{2}\frac{r^2}{h^2}$
• C. $\frac{3}{2}\frac{h^2}{r^3}$
• D. $\frac{2}{4}\frac{h^2}{r^2}$

Answer 1

The correct option is B $\frac{3}{2}\frac{r^2}{h^2}$

Field at the centre $B_1=\frac{{\mu }_0}{4\pi }\times \frac{2\pi in}{r}=\frac{{\mu }_0}{2}.\frac{ni}{r}$

Field at a distance $h$ from the centre

$B_2=\frac{{\mu }_0}{4\pi }.\frac{2\pi ni{r}^2}{(r^2+h^2{)}^{3/2}}$

$=\frac{{\mu }_0}{2}.\frac{ni{r}^2}{e^3{(1+\frac{h^2}{r^2})}^{3/2}}$

$=B_1(1+\frac{h^2}{r^2})$ (By binomial theorem)

Hence $B$ is less than $B$ by a fraction $=\frac{3}{2}\frac{h^2}{r^2}$