Question

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

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

Answer 1

Answer: (A)

The magnetic field normal to the plane of a coil of $N$ turns and radius $r$ having current $i$ on the axis at a distance $h$ from the centre is given by
$B_{\text {axis }}=\frac{\mu_{0} N i r^{2}}{2\left(r^{2}+h^{2}\right)^{\frac{3}{2}}}$
$=\frac{\mu_{0} N i r^{2}}{2\left(r^{2}\right)^{3 / 2}\left(1+\frac{h^{2}}{r^{2}}\right)^{\frac{3}{2}}}=\frac{\mu_{0} N i r^{2}\left[1+\frac{h^{2}}{r^{2}}\right]^{-\frac{3}{2}}}{2 r^{3}}$
[From Binomial theorem, $\left.(1+x)^{n}=1+n x\right]$
$B_{\text {axis }}=\frac{\mu_{0} N i}{2 r}\left(1-\frac{3 h^{2}}{2 r^{2}}\right)$ ......(i)
Magnetic field on the centre of a circular current carrying coil is given by
$B_{\text {centre }}=\frac{\mu_{0} N i}{2 r}$.....(ii)
From Eqs. (i) and (ii), we get
$B_{\text {axis }}=B_{\text {centre }}\left(1-\frac{3 h^{2}}{2 h^{2}}\right)$
$B_{\text {axis }}=B_{\text {centre }}-\frac{3 h^{2}}{2 r^{2}} \cdot B_{\text {entre }}$......(iii)
From Eq. (iii), it is clear that the magnetic field on
the axis is smaller than the field at the centre by a
fraction $\frac{3 h^{2}}{2 r^{2}}$