Question

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

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

Answer 1

Answer: (A)

$\frac{3}{2}\frac{h^2}{r^2}$

The magnetic field on the axis of a coil carrying current $I$, having $n$ turns, radius $r$ and at a distance $h$ from the centre of the coil, is given by

$B=\frac{{\mu }_0}{2\pi }\times \frac{2\pi NI{r}^2}{{(r^2+h^2)}^{3/2}}.........\left(1\right)$

The field at the centre is given by

$B_C=\frac{{\mu }_0}{4\pi }\times \frac{2\pi NI}{r}...........\left(2\right)$

$\therefore \frac{B}{B_C}=\frac{r^3}{{(r^2+h^2)}^{3/2}}=\frac{r^3}{{r^3(1+\frac{h^2}{r^2})}^{3/2}}=[1-\frac{3}{2}\frac{h^2}{r^2}]$

We have to find : $f=\frac{B_C-B}{B_C}=1-\frac{B}{B_C}=\frac{3}{2}\frac{h^2}{r^2}$