Question

If the ratio of magnitude of the magnetic field due to infinitesimally small segment of current carrying conductors $\text{A \& B}$ at a point $\text{P}$ is $3$. The value of current flowing in conductors $\text{A \& B}$ is $i\&2i$ respectively. Find the ratio of $r_B$ to $r_A$.

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\sqrt{6}$
• D. $\sqrt{\frac{1}{6}}$

Answer 1

The correct option is C $\sqrt{6}$


According to Biot -Savart law,

$B=\frac{{\mu }_{0}i}{4\pi }\frac{dl\sin \theta }{r^2}$

So, for both the conductors $\text{A \& B}$,

$B_A=\frac{{\mu }_0{i}_A}{4\pi }\frac{d{l}_A\sin {\theta }_A}{r_A^2}...\left(1\right)$

$B_B=\frac{{\mu }_0{i}_B}{4\pi }\frac{d{l}_B\sin {\theta }_B}{r_B^2}...\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$,

$\therefore \frac{B_A}{B_B}=\frac{\frac{{\mu }_0{i}_A}{4\pi }\frac{d{l}_A\sin \theta }{r_A^2}}{\frac{{\mu }_0{i}_B}{4\pi }\frac{d{l}_B\sin \theta }{r_B^2}}$

Here,
$i_A=i;i_B=2i;d{l}_A=d{l}_B=dl$
${\theta }_A={\theta }_B={90}^{\circ }$

$\frac{B_A}{B_B}=\frac{idl\sin {90}^{\circ }}{r_A^2}\frac{r_B^2}{2idl\sin {90}^{\circ }}=\frac{r_B^2}{2r_A^2}$

According to the problem, $B_A:B_B=3:1$

$\Rightarrow 3={\left(\frac{r_B}{r_A}\right)}^2\times \frac{1}{2}$

$\Rightarrow {\left(\frac{r_B}{r_A}\right)}^2=3\times 2$

$\therefore \frac{r_B}{r_A}=\sqrt{6}$

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