Question

A long wire carries a steady current of 1 Ampere. A ''S'' shaped conducting rod AB consisting of two semicircles each of radius r is placed in such a way that the centre C of the conducting wire is at a distance 2r from the end of the wire.The rod AB moves with velocity of 5$m{s}^{-1}$ along the direction of the current flow as shown in the figure. If the line joining the ends of the rod makes an angle ${60}^{\circ }$ with the wire then,

• A. the emf induced between the ends of the rod is (ln 4)$\mu V$
• B. the end A is at higher potential than end B.
• C. the end A is at lower potential than end B.
• D. the emf induced between the ends of the rod is (ln 3)$\mu V.$

Answer 1

Answer: (B), (D)

the end A is at higher potential than end B.
the emf induced between the ends of the rod is (ln 3)$\mu V.$

$AB=4r$
Induced emf in the conductor will be same as in AB.
Thus, we have a rod of effective length $l_{eff}=4r\cos \theta$ translating in a non-uniform magnetic field.
Consider a small element dx located at a distance x from the wire.

The magnetic field at this location is $B=\frac{{\mu }_{0}I}{2pix}$

The potential difference across the element is $dV=\frac{{\mu }_{0}I}{2\pi x}vdx$

$V={\int }_{d-2r\cos \theta }^{d+2r\cos \theta }dV={\int }_{d-2r\cos \theta }^{d+2r\cos \theta }\frac{{\mu }_{0}I}{2\pi x}vdx=\frac{{\mu }_{0}Iv}{2\pi }\ln \left(\frac{d+2r\cos \theta }{d-2r\cos \theta }\right)=2\times \frac{{\mu }_0}{4\pi }\times 1\times 5\left(\ln \frac{2r+2r\cos {30}^0}{2r-2r\cos {30}^0}\right)$

$\therefore V={10}^{-7}\times 10\times \frac{1+cos{30}^0}{1-\cos {30}^0}={10}^{-6}\times \ln \left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)=ln\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)\mu V$