Question

A copper rod of length $1\text{m}$ is moving at a uniform speed of $2.5\text{m/s}$ parallel to a long straight wire carrying a current of $2\text{A}$ as shown in the figure. Rod is perpendicular to the wire, with its ends at a distance of $1\text{m}$ and $2\text{m}$ from it. Calculate the emf induced $\left(\text{in volt}\right)$ in the rod.

• A. ${10}^{-6}\ln 2$
• B. ${10}^{-6}\ln 4$
• C. ${10}^{-6}\ln 5$
• D. ${10}^{-6}$

Answer 1

The correct option is A ${10}^{-6}\ln 2$

The magnetic field due to a long current carrying wire at a distance $x$ is,

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

Let us consider an element of length $dx$ in rod at a distance $x$ from the straight wire as shown in the figure.



Due to the motion of the element $dx$ in the rod in magnetic field $B$, the motional emf will be,

$dE=Bvdx$

$\therefore$ Total induced emf is given by,

$E=\int dE={\int }_a^{b}Bvdx$

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

Substituting the data given in question,

$E=\frac{4\times \pi {10}^{-7}\times 2\times 2.5}{2\pi }\ln \left(\frac{2}{1}\right)$

$\therefore E={10}^{-6}\ln \left(2\right)\text{V}$

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