Question

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

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

Answer 1

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

At the centre of coil the magnetic field is,
$B_1=\frac{{\mu }_{0}ni}{2r}$

On the axis of the coil the value of magnetic field is,

$B_2=\frac{n{\mu }_{0}i{r}^2}{2(r^2+h^2{)}^{3/2}}$

Fractional decrease in magnetic field will be:

$\frac{\Delta B}{B}=\frac{B_1-B_2}{B_1}$

$\frac{\Delta B}{B}=1-\frac{B_2}{B_1}$

$\frac{\Delta B}{B}=1-\frac{r^3}{(r^2+h^2)3/2}$

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

$\frac{\Delta B}{B}=[1-{\left(\frac{r^2+h^2}{r^2}\right)}^{\frac{-3}{2}}]$

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

Since $h<<r$, neglecting the higher order terms in the expansion, we can write

${(1+\frac{h^2}{r^2})}^{-3/2}=1-\frac{3}{2}\left(\frac{h^2}{r^2}\right)$

$\left[\right(1+nx{)}^{-1}=1-nx+\frac{n(n-1)}{2}{x}^2+.......]$

$\Rightarrow \frac{\Delta B}{B}=1-[1-\frac{3}{2}\left(\frac{h^2}{r^2}\right)]$

or, $\frac{\Delta B}{B}=\frac{3}{2}\left(\frac{h^2}{r^2}\right)$

$\therefore \frac{\Delta B}{B}=\frac{3h^2}{2r^2}$

Why this Question ? Note: The magnetic field due to a circular current carrying loop having $n$ turns on an axial point is given by, $B_{\text{axis}}=\frac{n{\mu }_{0}i{r}^2}{2(r^2+x^2{)}^{3/2}}$