Question

Two circular coils, $X$ and $Y$ have the equal number of turns and carry equal currents in the same sense and subtend some solid angle at point $O$ with their centers at the same line. The smaller coil $X$ is midway between point $O$ and coil $Y$. 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. $\frac{B_Y}{B_X}=2$
• B. $\frac{B_Y}{B_X}=\sqrt{2}$
• C. $\frac{B_Y}{B_X}=\frac{1}{\sqrt{2}}$
• D. $\frac{B_Y}{B_X}=\frac{1}{2}$

Answer 1

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


$△OBC$ and $△OAD$ are similar triangles.

$\Rightarrow \frac{BC}{AD}=\frac{OB}{OA}$

$\Rightarrow \frac{R_Y}{R_X}=\frac{2d}{d}$

$\Rightarrow R_Y=2R_X$

Let, $R_X=R\Rightarrow R_Y=2R$

Also, magnetic field due to coil $X$ at point $O$ on its axis,

$B_X=\frac{{\mu }_{o}NI{R}^2}{2(R^2+d^2{)}^{3/2}}$

Similarly, magnetic field due to coil $Y$ at point $O$ on its axis,

$B_Y=\frac{{\mu }_{o}NI(2R{)}^2}{2\left[\right(2R{)}^2+(2d{)}^2{]}^{3/2}}$

So,

$\frac{B_Y}{B_X}=\frac{4(R^2+d^2{)}^{3/2}}{(4R^2+4d^2{)}^{3/2}}=\frac{1}{4^{1/2}}$

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

Hence, option $\left(D\right)$ is the correct answer.