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: Understanding the Concept:
The figure shows a rectangular circuit with a resistor on the left and a rod of length \(L\) on the right moving with speed \(V\) in a field \(B\) pointing into the page. The flux through the loop grows, so an emf appears.

Step 2: Emf and current:
\[ \varepsilon = BLV, \qquad I = \frac{\varepsilon}{R} = \frac{BLV}{R} \]

Step 3: Magnetic force on the rod:
The current in the rod feels a force \(F = BIL\), directed opposite to the motion (Lenz's law).
\[ F = B\cdot\frac{BLV}{R}\cdot L = \frac{B^2L^2V}{R} \]

Step 4: Result:
To keep a constant speed the applied force must equal this magnetic force, option (B). Option (A) \(BVL/R\) is not a force, since it equals the current in amperes. Options (C) and (D) have an extra factor of \(V\), which does not fit a force.

Final Answer:
The applied force must match B I L = B^2 L^2 V / R. \[ \boxed{\text{(B) }\dfrac{B^2L^2V}{R}} \]