Question

$PQ$ is an infinite current-carrying conductor. $AB$ and $CD$ are smooth conducting rods on which a conductor $EF$ moves with constant velocity $V$ as shown in figure. The force needed to maintain constant speed of $EF$ is

• A. $\frac{1}{VR}{\left[\frac{{\mu }_{0}IV}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$
• B. ${\left[\frac{{\mu }_{0}IV}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2\frac{1}{VR}$
• C. ${\left[\frac{{\mu }_{0}IV}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2\frac{V}{R}$
• D. $\frac{V}{R}{\left[\frac{{\mu }_{0}IV}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$
• E. answer required

Answer 1

The correct option is A $\frac{1}{VR}{\left[\frac{{\mu }_{0}IV}{2\pi }\ln \left(\frac{b}{a}\right)\right]}^2$

Magnetic field at a distance $r$ from the wire $=B\left(r\right)=\frac{{\mu }_{0}I}{2\pi r}$

Hence emf induced between EF $={\int }_a^{b}B\left(r\right)vdr=\frac{{\mu }_{0}Iv}{2\pi }ln\left(\frac{b}{a}\right)$

Thus current in the loop $=i=\frac{{\mu }_{0}Iv}{2\pi R}ln\left(\frac{b}{a}\right)$

Net external force on the rod $={\int }_a^{b}B\left(r\right)idr=\frac{v}{R}\left[\frac{{\mu }_{0}I}{2\pi }ln\right(\frac{b}{a}){]}^2$

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