Question

A metal rod of length 'L' completes the circuit as shown. The area of the circuit is perpendicular to magnetic field 'B'. Total resistance of the circuit is 'R'. The force needed to move the rod in the direction as shown with constant speed 'V' is

• A. \(\frac{\text{BVL}}{\text{R}}\)
• B. \(\frac{\text{B}^2\text{L}^2\text{V}}{\text{R}}\)
• C. \(\frac{\text{B}^2\text{L}^2\text{V}^2}{\text{R}}\)
• D. \(\frac{\text{BLV}^2}{\text{R}}\)

Hint:

The rod's emf drives a current, and the magnetic force on that current opposes the motion.

Answer 1

The Correct Option is B

Step 1: Power balance:
Electrical power dissipated: $P = \dfrac{\varepsilon^2}{R} = \dfrac{B^2L^2V^2}{R}$.

Step 2: Mechanical power:
Mechanical power supplied $= FV$ at constant speed.

Step 3: Equate:
$FV = \dfrac{B^2L^2V^2}{R} \Rightarrow F = \dfrac{B^2L^2V}{R}$, option (B).

Final Answer:
The needed force is B^2 L^2 V / R. \[ \boxed{\text{(B) }\dfrac{B^2L^2V}{R}} \]