Question

Two circular coils $X$ and $Y$, having equal number of turns and carrying equal currents in the same sense, subtend same solid angle at point $O$. If the smaller coil $X$ is midway between $O$ and $Y$ and if we represent the magnetic field 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. $\frac{B_y}{B_x}=2$
• C. $\frac{B_y}{B_x}=\frac{1}{2}$
• D. $\frac{B_y}{B_x}=\frac{1}{4}$

Answer 1

The correct option is C $\frac{B_y}{B_x}=\frac{1}{2}$

Solid angle formed at any point = $\frac{S}{r^2}$, where $S$ is area of the surface and $r$ is the perpendicular distance.

$\frac{\pi r_y^2}{(2d{)}^2}=\frac{\pi r_x^2}{(d{)}^2}$
$\therefore r_y=2r_x$

Magnetic field due to any coil at distance $x$
$B=\frac{{\mu }_{o}2nIA}{4\pi (x^2+a^2{)}^{\frac{3}{2}}}$

Magnetic field due to coil y
$B_y=\frac{{\mu }_{o}2nI\pi (2r_x{)}^2}{4\pi \left[\right(2d{)}^2+\left(2r_x{)}^2\right){]}^{\frac{3}{2}}}$

Magnetic field due to coil x
$B_x=\frac{{\mu }_{o}2nI\pi (r_x{)}^2}{4\pi [d^2+r_x^2){]}^{\frac{3}{2}}}$

$\therefore \frac{B_y}{B_x}=\frac{1}{2}$