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Question #43936

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass $m = 4.0 \text{ kg}$ is attached to a horizontal spring of force constant $k = 100 \text{ N/m}$ . The block is mechanically coupled to an inductor such that the inductance varies with the block's displacement $x$ as $L(x) = L_0 + \alpha x$ , where $L_0 = 20 \text{ mH}$ and $\alpha = 0.8 \text{ H/m}$ . The unextended position of the spring is at $x = 0$ . At $t = 0$ , a constant current source is connected to the circuit, rapidly establishing a steady current $I_0 = 5.0 \text{ A}$ before the block has time to move. The current is maintained at $I_0$ thereafter. What is the difference between the total work done by the current source and the energy dissipated as heat in the circuit's resistance, during the time interval from when the block starts moving until it reaches its maximum extension for the first time?

Options

A.
$4.0 \text{ J}$
B.
$2.0 \text{ J}$
C.
$8.0 \text{ J}$
D.
$1.0 \text{ J}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Transient Current in RL Circuit|Effective Spring Constant|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of electromechanical energy conversion and the Work-Energy Theorem in coupled systems.

A common trap is assuming that the net electrical energy supplied (minus Joule heating) strictly equals the change in magnetic potential energy ( $\Delta U_m = \frac{1}{2} I_0^2 \Delta L$ ).

Students falling for this will incorrectly select 2.0 J.

However, they fail to account for the mechanical work done by the magnetic force ( $W_{mag} = \int F_m dx = \frac{1}{2} I_0^2 \Delta L$ ), which is ultimately stored as elastic potential energy in the spring.

For a constant current source, the total non-dissipative work done is $I_0^2 \Delta L$ , exactly twice the change in magnetic energy, because the source must supply energy for both the magnetic field and the mechanical work.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Transient Current in RL Circuit
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free