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Question #43302

Difficulty: 8/10 | 🟢 Free

The Question

A variable inductor has an inductance given by $L(x) = L_0 e^{-\alpha x}$ , where $x$ is the position of a movable magnetic core, $L_0$ is a positive constant in Henrys ( $\text{H}$ ), and $\alpha$ is a positive constant in $\text{m}^{-1}$ . The inductor is connected in series with a resistor of resistance $R$ ( $\Omega$ ) and a programmable voltage source that maintains a strictly constant current $I_0$ ( $\text{A}$ ) in the circuit. The core is pulled mechanically in the $+x$ direction at a constant speed $v$ ( $\text{m/s}$ ). Let $P_{ext}$ be the external mechanical power required to maintain the core's motion, and $P_{mag} = \frac{dU_B}{dt}$ be the rate of change of the magnetic energy stored in the inductor. Which of the following correctly gives the ordered pair $(P_{ext}, P_{mag})$ at any position $x$ ?

Options

A.
$\left( \frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x}, \; -\frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x} \right)$
B.
$\left( \alpha v L_0 I_0^2 e^{-\alpha x}, \; -\frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x} \right)$
C.
$\left( -\frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x}, \; -\frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x} \right)$
D.
$\left( \frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x}, \; \frac{1}{2} \alpha v L_0 I_0^2 e^{-\alpha x} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Lenz’s Law Energy Conservation|Self-Inductance of Solenoid|Power-Velocity Relationship

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the relationship between mechanical work, electrical power, and magnetic energy in a variable inductance system.

A common misconception is assuming that the mechanical power input directly equals the rate of change of magnetic energy ( $P_{ext} = P_{mag}$ ), or incorrectly applying the magnetic force formula $F_{mag} = -\frac{\partial U_B}{\partial x}$ which is only valid for constant flux, not constant current.

For a constant current system, $F_{mag} = +\frac{\partial U_B}{\partial x}$ .

Furthermore, students often neglect the motional EMF term $I\frac{dL}{dt}$ in Faraday's law.

By requiring the power balance $P_{ext} + I^2\frac{dL}{dt} = \frac{d}{dt}(\frac{1}{2}LI^2)$ , the problem forces the realization that the mechanical work done is exactly the negative of the change in magnetic energy, as the electrical source absorbs the difference.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Lenz’s Law Energy Conservation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free