Question

An infinite wire carries a current $I$. An $S$ shaped conducting rod of two semicircles each of radius $r$ is placed at angle $\theta$ to the horizontal as shown below. The centre of the conductor is at a distance $d$ from the wire. If rod translates parallel to the wire with velocity $u$, calculate the emf induced across the ends of the rod.

• A. $\frac{{\mu }_{0}iu}{2\pi }\ln \left(\frac{d+2r\cos \theta }{d-2r\cos \theta }\right)$
• B. $\frac{{\mu }_{0}iu}{4\pi }\ln \left(\frac{d+2r\cos \theta }{d-2r\cos \theta }\right)$
• C. $\frac{{\mu }_{0}iu}{4\pi }\ln \left(\frac{d+4r\cos \theta }{d}\right)$
• D. $\frac{{\mu }_{0}iu}{2\pi }\ln \left(\frac{d+4r\cos \theta }{d}\right)$

Answer 1

The correct option is A $\frac{{\mu }_{0}iu}{2\pi }\ln \left(\frac{d+2r\cos \theta }{d-2r\cos \theta }\right)$

The magnetic field at a distance $x$ due to a current carrying straight conductor is
$dB=\frac{{\mu }_{0}i}{2\pi x}$

$\therefore \text{emf}$ induced, $dE=dBvdx$

Now, net induced emf,

$E=\int dE=\int dBvdx$

$E=\int \frac{{\mu }_{0}i}{2\pi x}udx$

$E=\frac{{\mu }_{0}iu}{2\pi }\int \frac{dx}{x}$

$E=\frac{{\mu }_{0}iu}{2\pi }{\int }_{d-2r\cos \theta }^{d+2r\cos \theta }\frac{dx}{x}$

$\therefore E=\frac{{\mu }_{0}iu}{2\pi }\ln \left(\frac{d+2rcos\theta }{d-2rcos\theta }\right)$

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