Question

The distance at which the magnetic field on axis as
compared to the magnetic field atlthe centre of the coil
carrying current/ and radius R is 1/8, would be

• A. R
• B. $\sqrt 2 R$
• C. 2R
• D. $\sqrt 3 R$

Answer 1

Answer: (D)

If a coil of radius R is carrying current L, then magnetic field
on its axis at a distance x from its centre is given by
$ B_{axis}= \frac {\mu _0}{4 \pi} \frac {2 \pi IR^2}{(x^2+R^2)^{3/2}} ...(i) $
At centre
$ B_{centre}= \frac {\mu _0I}{2R}$
Dividing Eq. (i) by Eq. (ii), we get
$ \frac {B_{axis}}{B_{centre}}= \frac {\mu _0IR^2}{2(x^2+R^2)^{3/2}} \times \frac {2R}{\mu _0 I} $
$ = \frac {R^3}{(x^2+R^2)^{3/2}}= \frac {1}{8} $
or $\, \, \, \, \, \, \, \, \, \, \frac {R}{(x^2+R^2)^{1/2}}= \frac {1}{2}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x=\sqrt 3 R $