Question

Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sense and subtend same solid angle at point $O$. If the smaller coil, $X$ is midway between $O$ and $Y$, then if we represent the magnetic induction due to bigger coil $Y$ at $O$ as $B_Y$ and that due to smaller coil $X$ at $O$ as $B_X$ then

• A. $\frac{B_Y}{B_X}=1$
• B. $\frac{B_Y}{B_X}=\frac{1}{2}$
• C. $\frac{B_Y}{B_X}=\frac{1}{4}$
• D. $\frac{B_Y}{B_X}=2$

Answer 1

The correct option is B $\frac{B_Y}{B_X}=\frac{1}{2}$

Magnetic field on the axis of a circular coil,

$B=\frac{{\mu }_{0}ni{R}^2}{2(x^2+R^2{)}^{\frac{3}{2}}}$

Where,
$R$ is radius of coil, $x$ is distance of point from centre, $n$ is no. of turns and $i$ is current

Magnetic field at $O$ due to bigger coil $Y$ is

$B=\frac{{\mu }_0}{4\pi }\frac{2\pi (2r{)}^2}{[d^2+(2r{)}^2{]}^{\frac{3}{2}}}$

$B_Y=\frac{{\mu }_0}{4\pi }\frac{8\pi i{r}^2}{[d^2+4r^2{]}^{\frac{3}{2}}}$


Magnetic field at $O$ due to smaller coil $X$ is

$B_X=\frac{{\mu }_0}{4\pi }\frac{2\pi i(r{)}^2}{{[{\left(\frac{d}{2}\right)}^2+r^2]}^{\frac{3}{2}}}$

$B_X=\frac{{\mu }_0}{4\pi }\frac{16\pi i{r}^2}{[d^2+4r^2{)}^{\frac{3}{2}}}$

Taking ratio of $B_Y$ and $B_X$,

$\frac{B_Y}{B_X}=\frac{1}{2}$