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Question #81170

Difficulty: 5/10 | 🟢 Free

The Question

A horizontal cylindrical container is filled with an ideal gas and closed by a movable, rigid, conducting piston of radius $r$ and negligible mass. The space outside the cylinder is a vacuum. The piston acts as a single-turn circular coil of resistance $R$ . A constant ideal DC voltage source $V_0$ is connected across the coil. A radially outward magnetic field exists such that its radial component at the rim of the piston is $B = \alpha x$ , where $x$ is the distance of the piston from the closed end of the cylinder and $\alpha$ is a positive constant. The polarity of the voltage source is such that the current it drives produces a magnetic force on the piston pointing towards the closed end. Heat is slowly supplied to the gas such that the piston expands outwards at a constant velocity $v_0$ . Find the total work done by the gas as the piston moves from $x = L$ to $x = 2L$ . Assume all quantities are in standard SI units.

Options

A.
$\frac{3\pi r \alpha V_0 L^2}{R} + \frac{28\pi^2 r^2 \alpha^2 v_0 L^3}{3R}$
B.
$\frac{3\pi r \alpha V_0 L^2}{R} - \frac{28\pi^2 r^2 \alpha^2 v_0 L^3}{3R}$
C.
$\frac{3\pi r \alpha V_0 L^2}{R}$
D.
$\frac{2\pi r \alpha V_0 L^2}{R} + \frac{4\pi^2 r^2 \alpha^2 v_0 L^3}{R}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Cyclic Process Geometry|Force on Current-Carrying Loop|Ideal Gas Assumptions

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the coupling of thermodynamic work with electromagnetic damping in a non-uniform magnetic field.

The primary trap lies in the misapplication of Lenz's Law and the 'back-EMF' concept.

Students often blindly subtract the motional EMF from the source voltage ( $I = (V_0 - \varepsilon)/R$ ), assuming it must act as a back-EMF like in a standard motor.

However, here the gas does mechanical work to drive the piston outwards against the $V_0$ -induced force, making the system act as a generator.

The motional EMF therefore aids the external voltage in opposing the expansion, requiring the EMFs to be added.

Furthermore, the non-uniform field necessitates a differential work-energy integration rather than a simple average.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Cyclic Process Geometry
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free