Question

The field normal to the plane of a coil of $n$ turns and radius $r$ carrying a current $I$ is measured on the axis of the coil at a small horizontal distance $h$ from the centre of the coil. This is smaller than the field at the centre 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

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

The field at center is given by $B_c=\frac{{\mu }_{o}i}{2r}$
and on its axis at a distance h is given by $B_a=\frac{{\mu }_{o}i{r}^2}{2(r^2+h^2{)}^{3/2}}$ when $h<<r$

then $B_a=\frac{{\mu }_{o}i}{2R(1+(h/r{)}^2{)}^{3/2}}$
By applying binomial approximation we get,
$B_a=\frac{{\mu }_{o}i(1-\frac{3h^2}{2r^2})}{2r}$
Now Difference in the field at center and at distance h is $\Delta B=B_c-B_a$
$\Delta B=\frac{{\mu }_{o}i\times 3h^2}{4r^3}$
Now $\Delta B/B=\frac{3h^2}{2r^2}$
Hence option(A)