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Question #42993

Difficulty: 8/10 | 🟢 Free

The Question

Two rigid coils, 1 and 2, have self-inductances $L_1$ and $L_2$ . Their mutual inductance depends on the separation $x$ between them as $M(x) = M_0 \left(1 - \frac{x}{d}\right)$ for $0 \le x \le d$ , where $M_0$ and $d$ are positive constants. Coil 1 is held fixed at $x = 0$ . Coil 2 is moved slowly from $x = 0$ to $x = d$ by an external mechanical agent. During this displacement, the currents in the coils are maintained at constant positive values $I_1$ and $I_2$ respectively by ideal batteries. Let $W_{ext}$ be the mechanical work done by the external agent and $W_{elec}$ be the net extra electrical work done by the batteries (excluding the work done to compensate for Joule heating). Which of the following represents the pair $(W_{ext}, W_{elec})$ ?

Options

A.
$(M_0 I_1 I_2, -2 M_0 I_1 I_2)$
B.
$(-M_0 I_1 I_2, 0)$
C.
$(-M_0 I_1 I_2, 2 M_0 I_1 I_2)$
D.
$(M_0 I_1 I_2, -M_0 I_1 I_2)$

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

Energy Stored in Inductor|Mutual Inductance Reciprocity|Work-Energy Theorem Validity

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental thermodynamic distinction between constant-flux and constant-current electro-mechanical systems.

Students often incorrectly apply the work-energy theorem by equating the mechanical work done by the external agent directly to the change in magnetic potential energy ( $W_{ext} = \Delta U_m$ ), which yields $-M_0 I_1 I_2$ .

This assumes an isolated 'fixed-flux' system where no external electrical work is done.

However, because the currents are maintained constant by external batteries, the system is 'fixed-current'.

The batteries act as an energy pump, doing electrical work against the motional back-EMF generated by the changing mutual inductance.

The correct energy balance requires accounting for both mechanical and electrical work: $W_{ext} + W_{elec} = \Delta U_m$ , where the electrical work done by the sources is exactly twice the change in magnetic energy ( $W_{elec} = 2\Delta U_m$ ).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Energy Stored in Inductor
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free