Question

A rod of length $l$ is kept parallel to a long wire, carrying a constant current. It is moving away from the wire with a velocity $v$. Find the emf induced in the rod, at the instant when its distance from the long wire is $x$.

• A. $\frac{{\mu }_{o}ilv}{4\pi x}$
• B. $\frac{2{\mu }_{o}ilv}{\pi x}$
• C. $\frac{{\mu }_{o}ivx}{2\pi l}$
• D. $\frac{{\mu }_{o}ilv}{2\pi x}$

Answer 1

The correct option is D $\frac{{\mu }_{o}ilv}{2\pi x}$


The magnetic field due to a long current carrying wire is given by,

$B=\frac{{\mu }_{0}i}{2\pi x}$

Since, the rod is always parallel to the long wire, the magnetic field will be perpendicular and uniform throughout the rod.

The emf induced in the rod at the instant when its distance from the long wire is $x$,

$E=Bvl$

$=\frac{{\mu }_{0}i}{2\pi x}\times vl$

$\therefore E=\frac{{\mu }_{0}ilv}{2\pi x}$

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