Question

Two long parallel wires, having zero resistance, are connected to each other by a battery of $1.0\text{V}$. The separation between the wires is $0.5\text{m}$. A metallic bar of resistance $10\Omega$, placed perpendicular to the wires, moves on these wires. A uniform magnetic field of $0.02\text{T}$ is acting perpendicular to the plane containing the bar and the wires. Find the steady-state velocity of the bar.

• A. $10\text{m/s}$
• B. $150\text{m/s}$
• C. $50\text{m/s}$
• D. $100\text{m/s}$

Answer 1

The correct option is D $100\text{m/s}$

The initial current flowing through the bar, when the bar is at rest, is,

$i=\frac{E}{R}=\frac{1}{10}=0.1\text{A}$

The current carrying bar is placed in the magnetic field $B=0.02\text{T}$ perpendicular to the plane of paper and directed inwards.

The magnetic force acting on the bar is,

$F=BIl=0.02\times 0.1\times 0.5={10}^{-3}\text{N}$

Due to this force, the bar will start moving in the magnetic field.

If the instantaneous velocity of the bar is $v$, then the emf induced in the bar is,

$E=Blv=0.01v$

By the Lenz's law, $E$ will oppose the motion of bar, which will ultimately attain a steady velocity.

In this state, the induced emf will be equal to the applied voltage,

$\therefore 0.01v=1$

$\therefore v=100\text{m/s}$

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