Question

As shown in the figure, a rod is moving with velocity, $v=20\text{m/s}$. For the rod to move with constant velocity, we must need to apply an external force in the right direction. What is the value of this external force?

Given: Magnetic field, $B=3\text{T}$ and length of the rod, $l=1\text{m}$.

Assume resistance of the rod and connecting wires to be small.

• A. $30\text{N}$
• B. $60\text{N}$
• C. $45\text{N}$
• D. $90\text{N}$

Answer 1

The correct option is B $60\text{N}$


The resistors are forming a Wheatstone bridge.

The resultant resistance between points $P$ and $O$,

$R_{PO}=\frac{6\times 6}{6+6}=3\Omega$

Magnitude of external force required,

$F=\frac{B^2{l}^{2}v}{R_{PO}}=\frac{3^2\times 1^2\times 20}{3}=60\text{N}$

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