Question

A metal rod completes a circuit as shown in the figure. The circuit is normal to magnetic field $B=0.15\text{T}$. If the resistance of the circuit is $3\Omega$, then the force required to move the rod with a constant velocity of $2\text{m/s}$ will be

• A. $3.75\times {10}^{-3}\text{N}$
• B. $3.75\times {10}^{-2}\text{N}$
• C. $3.75\times {10}^{-5}\text{N}$
• D. $3.75\times {10}^{-4}\text{N}$

Answer 1

The correct option is A $3.75\times {10}^{-3}\text{N}$

Induced emf in the rod,

$\therefore E=Blv$

Magnitude of current in the circuit,

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

When a current carrying conductor moves in a magnetic field then, the force experienced by that conductor is given by

$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})$

$\Rightarrow F=Bil\sin \theta$

Here, $\theta ={90}^{\circ }$

$\therefore F=Bil=B\left(\frac{Blv}{R}\right)l=\frac{B^2{l}^{2}v}{R}$

$\Rightarrow F=\frac{{0.15}^2\times {0.5}^2\times 2}{3}$

$\Rightarrow F=3.75\times {10}^{-3}\text{N}$

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