Question

A rod of length $3\text{m}$ is kept parallel to a long wire carrying constant current $0.5\text{A}.$ It is moving away from the wire with a velocity of $10\text{m/s}.$ Find the induced emf in the rod when its distance from the long wire is $0.15\text{m}.$

• A. $3\times {10}^{-3}\text{V}$
• B. $2\times {10}^{-6}\text{V}$
• C. $5\times {10}^{-4}\text{V}$
• D. $2\times {10}^{-5}\text{V}$

Answer 1

The correct option is D $2\times {10}^{-5}\text{V}$

The magnetic field due to long wire at distance $x$ is given by,

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


$\therefore$ The emf induced in the rod is,

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

$=\frac{4\pi \times {10}^{-7}\times 0.5}{2\pi \times 0.15}\times 10\times 3$

$=2\times {10}^{-5}\text{V}$

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