Question

A wire is sliding on two parallel conducting rails placed at a separation of $1\text{m}$ as shown in figure. A uniform magnetic field of $2\text{T}$ exists in a direction perpendicular to rails. What force $F$ is necessary to keep the wire moving with a constant velocity of $1\text{cm/s}$?

• A. $8\times {10}^{-3}\text{N}$
• B. $8\times {10}^{-6}\text{N}$
• C. $4\times {10}^{-6}\text{N}$
• D. $8\times {10}^{-4}\text{N}$

Answer 1

The correct option is D $8\times {10}^{-4}\text{N}$

We know that, emf induced,

$E=Blv$

$\therefore$ Current in the circuit,

$i=\frac{Blv}{R}$

$\Rightarrow i=\frac{2\times 1\times 1\times {10}^{-2}}{50}=4\times {10}^{-4}\text{A}$

Now, force on the wire is,

$F=iBl$

$\Rightarrow F=4\times {10}^{-4}\times 2\times 1$

$\Rightarrow F=8\times {10}^{-4}\text{N}$

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