Question

A rod $PQ$ of length $L$ moves with a uniform velocity $v$ parallel to a long straight wire carrying a current $i$. The end $P$ of the rod always remains at a distance $r$ from the wire. Calculate the emf induced across the rod.

Take, $v=5.0\text{m/s},i=100\text{A},r=1.0\text{cm}$ and $L=19\text{cm},{\log }_{e}20=3$

• A. $7\times {10}^{-4}\text{V}$
• B. $5\times {10}^{-4}\text{V}$
• C. $3\times {10}^{-4}\text{V}$
• D. $1\times {10}^{-4}\text{V}$

Answer 1

The correct option is C $3\times {10}^{-4}\text{V}$

The rod $PQ$ is moving in the magnetic field, produced by the long, straight current-carrying wire.

The field is not uniform throughout the length of the rod (changing with distance).

Let us consider a small element $dx$ at distance $x$ from wire.


If magnetic field at the position of $dx$ is $B$ then induced emf in this element is,

$dE=Bv\left(dx\right)=\frac{{\mu }_{0}i}{2\pi x}v\left(dx\right)$

Now, induced emf in the entire length of the rod $PQ$ is,

$E=\int dE={\int }_{x_P}^{x_Q}\frac{{\mu }_0}{2\pi }\frac{i}{x}v\left(dx\right)$

At $P$, $x_P=r$ and at $Q$, $x_Q=r+L$.

Hence,

$E=\frac{{\mu }_{0}iv}{2\pi }{\int }_r^{r+L}\frac{dx}{x}$


$=\frac{{\mu }_{0}iv}{2\pi }[{\log }_{e}x{]}_r^{r+L}$


$=\frac{{\mu }_{0}iv}{2\pi }\left[{\log }_e\right(r+L)-{\log }_{e}r]$

$=\frac{{\mu }_{0}iv}{2\pi }\log (1+\frac{L}{r})$

Putting all the given values,

$E=(2\times {10}^{-7})\left(100\right)\left(5.0\right){\log }_e(1+\frac{19}{1})$

$=3\times {10}^{-4}\text{V}$

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