Question

A metal rod $\text{PQ}$ of length $l$ slides with a velocity $v$ on two parallel rails $\text{AB}$ and $\text{CD}$ parallel to a long straight wire $\text{XY}$ carrying a current $I$ as shown in the figure. A resistance $R$ is connected between the rails as shown. The velocity of rod $\text{PQ}$ is kept constant by applying force. Find the current induced in resistance $R.$

• A. $\frac{{\mu }_{0}Iv}{2\pi R}\ln \left(\frac{a+l}{a}\right)$
• B. $\frac{{\mu }_{0}Iv}{4\pi R}\ln \left(\frac{a+l}{a}\right)$
• C. $\frac{Blv}{R}$
• D. ${\mu }_{0}Iv\ln \left(\frac{a+l}{a}\right)$

Answer 1

The correct option is A $\frac{{\mu }_{0}Iv}{2\pi R}\ln \left(\frac{a+l}{a}\right)$

In this case emf is induced due to the change in magnetic flux which is due to the change in the area of $\text{ACPQ}$ with time.


The induced emf through an infinitesimally small area element of width $dx$ at a distance $x$ from $\text{XY}$ is given by

$dE=Bvdx=\frac{{\mu }_{0}I}{2\pi x}vdx$

So emf induced through entire rod will be,

$E={\int }_a^{a+l}\frac{{\mu }_{0}Iv}{2\pi }\frac{dx}{x}==\frac{{\mu }_{0}Iv}{2\pi }{\int }_a^{a+l}\frac{dx}{x}$

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

$\Rightarrow$ Current induced in $R$ is

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

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