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
$B_{\text{axis }}=\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2+h^2\right)}^{\frac{3}{2}}}$
$=\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2\right)}^{3/2}{\left(1+\frac{h^2}{r^2}\right)}^{\frac{3}{2}}}=\frac{{\mu }_{0}Ni{r}^2{\left[1+\frac{h^2}{r^2}\right]}^{-\frac{3}{2}}}{2r^3}$
[From Binomial theorem, $(1+x{)}^n=1+nx]$
$B_{\text{axis }}=\frac{{\mu }_{0}Ni}{2r}\left(1-\frac{3h^2}{2r^2}\right)$ ......(i)
Magnetic field on the centre of a circular current carrying coil is given by
$B_{\text{centre }}=\frac{{\mu }_{0}Ni}{2r}$.....(ii)
From Eqs. (i) and (ii), we get
$B_{\text{axis }}=B_{\text{centre }}\left(1-\frac{3h^2}{2h^2}\right)$
$B_{\text{axis }}=B_{\text{centre }}-\frac{3h^2}{2r^2}\cdot B_{\text{entre }}$......(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 $\frac{3h^2}{2r^2}$