Question

Two circular coils $X$ and $Y$ have equal number of turns and carry equal currents in the same sense and subtend the same solid angle at point $\left( O \right)$ . If the smaller coil $X$ is midway between $O$ and $Y$ , then if we represent 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. $\dfrac{{{B_Y}}}{{{B_X}}} = 1$
B. $\dfrac{{{B_Y}}}{{{B_X}}} = 2$
C. $\dfrac{{{B_Y}}}{{{B_X}}} = \dfrac{1}{2}$
D. $\dfrac{{{B_Y}}}{{{B_X}}} = \dfrac{1}{4}$

Answer 1

In this question, we are given two coils carrying the same current and having the same number of turns. As they subtend the same solid angle at point $O$ and the smaller coil is midway between the larger coil and point $O$ . By using geometry we can say, Radius of the larger coil is twice the radius of smaller coil and we can find out the magnetic field due to coil at a distance $x$ using the formula $B = \dfrac{{{\mu _0}NI{R^2}}}{{2{{\left( {{R^2} + {x^2}} \right)}^{3/2}}}}$
Magnetic field due at the axis of a circular current-carrying coil, $B = \dfrac{{{\mu _0}NI{R^2}}}{{2{{\left( {{R^2} + {x^2}} \right)}^{3/2}}}}$
Where $B$ is the magnetic field at the axis
$R$ is the radius of the coil
$I$ is the current carried by the coil
$N$ is the number of turns in the coil
$x$ is the distance from the center of the coil.

Complete step by step answer:
We are given two coils $X$ and $Y$ having an equal number of turns $N$ and carrying an equal amount of current $I$
Distance of centre of coil $X$ from the point $O$ is $d$
As coil $X$ is midway between $Y$ and $O$ ,
Distance of centre of coil $Y$ and $O$ is $2d$
Let radius of coil $Y$ be $R$ and radius of coil $X$ be $r$
Both the coils subtend same angle at point $O$ thus,
$\dfrac{r}{d} = \dfrac{R}{{2d}}$
$\Rightarrow R = 2r$
Now, the magnetic field at point $O$ due to bigger coil $Y$ , ${B_Y} = \dfrac{{{\mu _0}NI{R^2}}}{{2{{\left( {{R^2} + {{\left( {2d} \right)}^2}} \right)}^{3/2}}}}$
Substituting $R = 2r$ we get,
$\Rightarrow {B_Y} = \dfrac{{{\mu _0}NI\left( {4{r^2}} \right)}}{{2{{\left( {4{r^2} + 4{d^2}} \right)}^{3/2}}}}$
$\Rightarrow {B_Y} = \dfrac{{{\mu _0}NI\left( {4{r^2}} \right)}}{{2 \times 8{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}$
$\Rightarrow {B_Y} = \dfrac{{{\mu _0}NI{r^2}}}{{4{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}$
Magnetic field at point $O$ due to smaller coil $X$ , ${B_X} = \dfrac{{{\mu _0}NI{r^2}}}{{2{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}$
Therefore, $\dfrac{{{B_Y}}}{{{B_X}}} = \dfrac{{\dfrac{{{\mu _0}NI{r^2}}}{{4{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}}}{{\dfrac{{{\mu _0}NI{r^2}}}{{2{{\left( {{r^2} + {d^2}} \right)}^{3/2}}}}}} \Rightarrow \dfrac{1}{2}$
$\dfrac{{{B_Y}}}{{{B_X}}} = \dfrac{1}{2}$
Hence, the correct option is option C.

Note:
Magnetic field at the center of the coil is maximum and is equal to ${B_X} = \dfrac{{{\mu _0}NI}}{{2R}}$ . The magnetic field along the axis of the coil keeps on decreasing as we move away from the center of the coil and at far points, $x \gg R$ magnetic field becomes, $B = \dfrac{{{\mu _0}NI{R^2}}}{{2{x^3}}}$ .