Question

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

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

Answer 1

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

The emf induced across the rod is,

$E=Bvl=0.15\times 2\times 0.5=0.15\text{V}$

$\therefore$ Induced current in the rod will be,

$i=\frac{E}{R}=\frac{0.15}{3}=0.05\text{A}$

Now, force due to the magnetic field is,

$F=iBl=0.05\times 0.15\times 0.5$

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

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