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
$B_{axis}=\frac{{\mu }_0}{4\pi }\frac{2\pi I{R}^2}{(x^2+R^2{)}^{3/2}}...(i)$
At centre
$B_{centre}=\frac{{\mu }_{0}I}{2R}$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{B_{axis}}{B_{centre}}=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}\times \frac{2R}{{\mu }_{0}I}$
$=\frac{R^3}{(x^2+R^2{)}^{3/2}}=\frac{1}{8}$
or $\frac{R}{(x^2+R^2{)}^{1/2}}=\frac{1}{2}$
$\Rightarrow x=\sqrt{3}R$