Question

Consider an infinitely long current carrying cylindrical straight wire

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

Answer 1

The correct option is B

$\frac{2}{3}$

Step 1. Given Data,

Radius of cylindrical straight wire $=a$.

Distance from point $A=$ $\frac{a}{3}$

Distance from point $B=$ $2a$

We have to find the ratio of the magnetic field due to wire at distance $\frac{a}{3}$ and $2a$, respectively, from axis of wire.

Step 2. Find the ratio of magnetic field due to wire at distance $\frac{a}{3}$ and $2a$, respectively, from axis of wire,

Magnetic Field on point $A$

$B_A=\frac{{\mu }_{0}ir}{2{\mathrm{\pi a}}^2}$

Where, ${\mu }_0=4\mathrm{\pi }\times {10}^{-7}{\mathrm{Hm}}^{-1}$is a vacuum permiability, $i$ is current, $r$ is distance from one point and $a$ is also distance but from another point.)

$$\begin{gathered} \Rightarrow B_A=\frac{{\displaystyle \frac{{\mu }_{0}ia}{3}}}{2{\mathrm{\pi a}}^2} \\ \Rightarrow B_A=\frac{{\displaystyle {\mu }_{0}i}}{6\mathrm{\pi a}} \end{gathered}$$

Magnetic Field on point $B$

$$\begin{gathered} B_B=\frac{{\mu }_{0}ir}{2{\mathrm{\pi a}}^2} \\ \Rightarrow B_A=\frac{{\displaystyle {\mu }_{0}i\left(2a\right)}}{2\pi {\left(2a\right)}^2} \\ \Rightarrow B_A=\frac{{\displaystyle {\mu }_{0}i}}{4\mathrm{\pi a}} \end{gathered}$$

Step 3. Ratio of $\frac{B_A}{B_B}$.

$$\begin{gathered} \frac{B_A}{B_B}=\frac{{\displaystyle \frac{{\mu }_{0I}}{6\mathrm{\pi a}}}}{{\displaystyle \frac{{\mu }_{0I}}{4\mathrm{\pi a}}}} \\ \Rightarrow \frac{B_A}{B_B}=\frac{2}{3} \end{gathered}$$

Hence, the correct option is $\text{'}B\text{'}$.