Question

$\text{PQ}$ is an infinite current carrying conductor. $\text{AB}$ and $\text{CD}$ are smooth conducting rods on which a conductor $\text{EF}$ moves with constant velocity $v$ as shown. The force needed to maintain constant speed of $\text{EF}$ is

• A. $\frac{v}{R}{\left[\frac{{\mu }_{0}I}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$
• B. $\frac{2v}{R}{\left[\frac{{\mu }_{0}I}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$
• C. $\frac{v}{R}{\left[\frac{{\mu }_{0}I}{2\pi }\ln 2\right]}^2$
• D. $0$

Answer 1

The correct option is A $\frac{v}{R}{\left[\frac{{\mu }_{0}I}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$

The magnetic field at a distance $r$ from the infinite wire is,

$B=\frac{{\mu }_{0}I}{2\pi r}$


Hence, emf induced across $EF$

$E={\int }_a^{b}Bvdr=\frac{{\mu }_{0}Iv}{2\pi }{\int }_a^b\frac{dr}{r}$

$\therefore E=\frac{{\mu }_{0}Iv}{2\pi }\ln \left(\frac{b}{a}\right)$

Thus, induced current in the loop,

$i=\frac{E}{R}=\frac{{\mu }_{0}Iv}{2\pi R}\ln \left(\frac{b}{a}\right)$

External force on the element $dr$,

$\therefore dF=Bidr$

So, net external force on the conductor $\text{EF}$ is

$F={\int }_a^{b}Bidr=i{\int }_a^{b}Bdr$

$=\frac{{\mu }_{0}Iv}{2\pi R}\ln \left(\frac{b}{a}\right)\times \frac{{\mu }_{0}I}{2\pi }{\int }_a^b\frac{dr}{r}$

$\therefore F=\frac{v}{R}{\left[\frac{{\mu }_{0}I}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$

Hence, same amount of force must be applied opposite to this external force to maintain a constant speed.

Hence, option $\left(\text{A}\right)$ is the correct answer.