Question

A long wire carries a steady current of $1A$. 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 $5m{s}^{-1}$ along the direction of the current flow as shown in the figure. If the line joining the ends of the rod makes an angle ${30}^{\circ }$ 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 correct options are
B the end A is at higher potential than end B.
D the emf induced between the ends of the rod is ($\ln 3)\mu V.$

Joining the ends A and B, component of length of the rod, perpendicular to velocity v is AB$sin\theta =4rsin\theta$

Consider an element of length $dx$ at a distance $x$ from the wire, magnetic field at this element is $B=\frac{{\mu }_{0}I}{2\pi x}.$ Hence potential difference induced across this element is

$de=-\frac{{\mu }_{0}I}{2\pi x}vdx$

$\therefore$ P.D. induced between A and B is,

$e=-\int de=-{\int }_{2r-2rsin\theta }^{2r+2rsin\theta }\frac{{\mu }_{0}Iv}{2\pi x}dx=-\frac{{\mu }_{0}Iv}{2\pi }{\left[lnx\right]}_{2r(1-sin\theta )}^{2r(1+sin\theta )}$
$=-\frac{{\mu }_{0}Iv}{2\pi }ln\left(\frac{1+sin\theta }{1-sin\theta }\right)=-\frac{4\pi \times {10}^{-7}\times 1\times 5}{2\pi }ln3$
$=-{10}^{-6}ln3V=-ln3\mu V$
the direction of EMF induced can be found from equation $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$
assuming z axis out of the plane of paper and y axis along length of current and x axis perpendicular to length of wire, forming right hand coordinate system
$\overrightarrow{v}=v\hat{j};\overrightarrow{B}=B-\hat{k}$
$\overrightarrow{F}=q(v\hat{j}\times B-\hat{k})$
$\overrightarrow{F}=qvB(-\hat{i})$
The positive charges will shift towards point A, thus point A will be at higher potential.