Question

A rod of length $l=0.5\text{m}$ is rotating with an angular speed $\omega =10\text{rad/s}$ about its one end which is at a distance $a=1\text{m}$ from an infinitely long wire carrying current $i=1\text{A}.$ Find the emf induced in the rod at the instant shown in the figure. Rod and wire both are lying in the same plane.

• A. $4\times {10}^{-6}[1-\ln (2\left)\right]\text{V}$
• B. $0\text{V}$
• C. $2\times {10}^{-6}[0.5-\ln (1.5\left)\right]\text{V}$
• D. $2\times {10}^{-6}[2-\ln (3\left)\right]\text{V}$

Answer 1

The correct option is C $2\times {10}^{-6}[0.5-\ln (1.5\left)\right]\text{V}$

Consider a small segment of rod of length $dx$ at a distance $x$ from one end of the rod.


Emf induced in the segment is,

$dE=B_{x}vl=\frac{{\mu }_{0}i}{2\pi (x+a)}\left(x\omega \right)dx$

$\therefore$ Emf induced across the whole rod at the given instant is,

$E=\int dE={\int }_0^l\frac{{\mu }_{0}i}{2\pi (x+a)}\left(x\omega \right)dx$

$E=\frac{{\mu }_{0}i\omega }{2\pi }{\int }_0^l(1-\frac{a}{x+a})dx$

$\therefore E=\frac{{\mu }_{0}i\omega }{2\pi }[l-a\ln \left(\frac{l+a}{a}\right)]$

From the data given in the question,

$E=\frac{4\pi \times {10}^{-7}\times 1\times 10}{2\pi }[0.5-1\times \ln \left(\frac{0.5+1}{1}\right)]$

$\therefore E=2\times {10}^{-6}[0.5-\ln (1.5\left)\right]\text{V}$

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