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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The fields at axis and centre respectively are given by
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 {πir}^{2}}{{(r^{2} + h^{2})}^{3 / 2}}$
and $B_{centre} = \frac{μ_{0}}{4 π} \times \frac{2 πi}{r}$
$∴ \frac{B_{axis}}{B_{centre}} = \frac{r^{3}}{{(r^{2} + h^{2})}^{3 / 2}} = \frac{r^{3}}{r^{3} {[1 + \frac{h^{2}}{r^{2}}]}^{3 / 2}}$
$= {[1 + \frac{h^{2}}{r^{2}}]}^{- 3 / 2} = (1 - \frac{3 h^{2}}{2 r^{2}})$
or $(1 - \frac{B_{axis}}{B_{centre}}) = \frac{3}{2} \frac{h^{2}}{r^{2}}$
or $\frac{B_{centre} - B_{axis}}{B_{centre}} = \frac{3}{2} \frac{h^{2}}{r^{2}}$