Question

A rod $\text{AB}$ of length $0.5\text{m}$ can slide freely on a pair of smooth, horizontal rails as shown in the figure. A magnetic field $B_0$ exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the left end by a combination of a resistor and battery, as shown in the figure. The constant speed with which the rod slides after applying the external force, $F=0.5\text{N}$ must be:

• A. $4{\text{ms}}^{-1}$
• B. $16{\text{ms}}^{-1}$
• C. $20{\text{ms}}^{-1}$
• D. $8{\text{ms}}^{-1}$

Answer 1

The correct option is B $16{\text{ms}}^{-1}$

Let, $I$ be the current in the circuit from $\text{A}$ to $\text{B}$.

Magnitude of magnetic force on the rod $\text{AB}$,

$F_m=|I(\overrightarrow{l}\times {\overrightarrow{B}}_0)|=IlB\sin \theta$

$\because \theta ={90}^{\circ },B_0=0.5\text{T}\text{and}l=0.5\text{m}$

$\therefore F_m=I\times 0.5\times 0.5=0.25I$


$F_m$ acts on the rod $\text{AB}$ towards right.

To keep the rod move with constant velocity, a force $F=0.5\text{N}$ is applied on the rod $\text{AB}$ towards left.

Equating $F_m$ with $F$,

$0.25I=0.5$

$\Rightarrow I=2\text{A}$

Applying $\text{KVL}$ to the circuit.

$V_A-V_B=8-IR$

$=8-(2\times 2)=4\text{V}[\because R=2\Omega ]$

As, if $v$ is the constant velocity with which the rod is moving , then the emf induced across the rod,

$E=V_A-V_B=B_{0}lv$

$\Rightarrow V_A-V_B=0.5\times 0.5v=0.25v$

$\Rightarrow 4=0.25v$

$\therefore v=16{\text{ms}}^{-1}$

Hence, $\left(\text{B}\right)$ is the correct answer.