Question

The figure shows two coaxial circular loops $1$ and $2$ carrying equal current $I$, form the same solid angle $\theta$ at point $O$. If $B_1$ and $B_2$ are the magnetic fields produced at point $O$ due to loop $1$ and $2$ respectively, then,

• A. $\frac{B_1}{B_2}=1$
• B. $\frac{B_1}{B_2}=2$
• C. $\frac{B_1}{B_2}=8$
• D. $\frac{B_1}{B_2}=4$

Answer 1

The correct option is B $\frac{B_1}{B_2}=2$

Considering the semi-vertex angle of the cone shown in the diagram, we get

$\tan \frac{\theta }{2}=\frac{r_1}{x}=\frac{r_2}{2x}$

$\therefore r_2=2r_1$

Let, $r_1=R$ then $r_2=2R$

Now, magnetic field at point $O$,

$\left(a\right)$ due to loop $1$:

$B_1=\frac{{\mu }_{o}I{R}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}....\left(i\right)$

$\left(b\right)$ due to loop $2$:

$B_2=\frac{{\mu }_{o}I(2R{)}^2}{2\left[\right(2R{)}^2+(2x{)}^2{]}^{\frac{3}{2}}}$

$=\frac{2{\mu }_{o}I{R}^2}{[4R^2+4x^2{]}^{\frac{3}{2}}}$

$\Rightarrow B_2=\frac{{\mu }_{o}I{R}^2}{4[R^2+x^2{]}^{\frac{3}{2}}}.....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{B_1}{B_2}=2$

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