Question

The power required to push the arm of a rectangular conductor with a constant speed \(v\) in a motor is directly proportional to

• A. \(v\)
• B. \(v^2\)
• C. \(\sqrt{v}\)
• D. \(\sqrt[3]{v}\)
• E. \(v^3\)

Hint:

In motional emf problems, power often depends on velocity squared because both induced current and force depend on velocity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a conductor moves through a magnetic field, a motional electromotive force (EMF) is induced across its ends. If this conductor is part of a closed circuit, the EMF drives a current. This current, flowing through the conductor which is still in the magnetic field, will experience a magnetic force (Lorentz force). According to Lenz's law, this force will oppose the motion. To maintain a constant speed \(v\), an external agent must apply a force equal in magnitude to this opposing magnetic force. The power delivered by this external agent is what the question asks for.
Step 2: Key Formula or Approach:
1. Calculate the induced motional EMF: \( \mathcal{E} = BLv \), where B is the magnetic field, L is the length of the arm, and v is its speed.
2. Calculate the current in the circuit: \( I = \frac{\mathcal{E}}{R} \), where R is the resistance of the circuit.
3. Calculate the magnetic drag force on the arm: \( F_{mag} = ILB \).
4. To move at a constant speed, the applied force must equal the magnetic force: \( F_{app} = F_{mag} \).
5. Calculate the power delivered by the applied force: \( P = F_{app} \cdot v \).
Step 3: Detailed Explanation:
1. The induced EMF in the moving arm is \( \mathcal{E} = BLv \).
2. The current flowing through the arm is \( I = \frac{\mathcal{E}}{R} = \frac{BLv}{R} \).
3. The magnetic force opposing the motion is \( F_{mag} = ILB = \left(\frac{BLv}{R}\right)LB = \frac{B^2L^2v}{R} \).
4. The external force required to maintain constant speed is \( F_{app} = F_{mag} = \frac{B^2L^2v}{R} \).
5. The power required is the rate at which this force does work: \( P = F_{app} \times v \).
\[ P = \left(\frac{B^2L^2v}{R}\right) \times v = \frac{B^2L^2v^2}{R} \] In this expression, the magnetic field B, the length of the arm L, and the resistance R are all constants for a given setup. Therefore, the power P is directly proportional to the square of the speed v.
\[ P \propto v^2 \] Step 4: Final Answer:
The power required is directly proportional to \( v^2 \). This corresponds to option (B).