Question

Two circular coils $X$ and $Y$ have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point $\left(O\right)$. 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 $B,$ 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 $\frac{B_y}{B_x}=\frac{1}{2}$

We know magnetic induction due to a circular currrent carrying coil on its axis at distance x from its center is given by,
$B=\frac{{\mu }_{o}nI{R}^2}{2{(x^2+R^2)}^{3/2}}$
Now Let larger coil be at a distance $x$ and have radius $R$. then smaller coil will be have radius $R/2$ and at a distance $x/2$ since it is midway and subtends same angle as subtended by larger coil at point O.
Magnetic field due to larger coil at O
$B_y=\frac{{\mu }_{o}nI{R}^2}{2{(x^2+R^2)}^{3/2}}$
and due to smaller coil will be
$B_x=\frac{{\mu }_{o}nI{\left(R/2\right)}^2}{2{({\left(x/2\right)}^2+{\left(R/2\right)}^2)}^{3/2}}$
$B_x=\frac{{\mu }_{o}nI{\left(R\right)}^2}{4{({\left(x\right)}^2+{\left(R\right)}^2)}^{3/2}}$
Therefore, $B_y/B_x=1/2$
option(C)