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Question #43299

Difficulty: 8/10 | 🟢 Free

The Question

A square conducting loop of mass $m$ , side length $L$ , and electrical resistance $R$ is released from rest at $y=0$ and falls vertically downward in the $xy$ -plane. The loop falls through a viscous fluid that exerts a drag force $\vec{F}_v = -k\vec{v}$ , where $k$ is a positive constant and $\vec{v}$ is the velocity of the loop. A non-uniform horizontal magnetic field exists in the region, given by $\vec{B} = \beta y \hat{k}$ , where $\beta$ is a positive constant and $\hat{k}$ is perpendicular to the plane of the loop. Assuming the loop remains perfectly vertical and reaches its terminal velocity before falling a total distance $H$ , what is the total Joule heat dissipated in the loop during this fall? (Acceleration due to gravity is $g$ in the $+y$ direction. Neglect the self-inductance of the loop).

Options

A.
$\frac{\beta^2 L^4}{\beta^2 L^4 + k R} \left( mgH - \frac{m^3 g^2 R^2}{2(\beta^2 L^4 + k R)^2} \right)$
B.
$\frac{\beta^2 L^4}{\beta^2 L^4 + k R} mgH$
C.
$mgH - \frac{m^3 g^2 R^2}{2(\beta^2 L^4 + k R)^2}$
D.
$\frac{k R}{\beta^2 L^4 + k R} \left( mgH - \frac{m^3 g^2 R^2}{2(\beta^2 L^4 + k R)^2} \right)$

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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

Lenz’s Law Energy Conservation|Work Done by Friction|Mechanical Energy Conservation

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of electromagnetic induction, fluid mechanics, and the Work-Energy Theorem.

A primary trap is the non-uniform magnetic field; students often incorrectly assume this leads to a position-dependent magnetic drag, failing to realize that the spatial derivative of the flux yields a constant effective damping coefficient.

The most lethal trap, however, lies in the energy conservation phase.

Students frequently equate the total loss in gravitational potential energy ( $mgH$ ) directly to Joule heating, ignoring both the kinetic energy retained by the loop at terminal velocity and the work done by the viscous fluid.

By proving that the instantaneous power dissipated by both non-conservative forces scales with $v^2$ , students can bypass complex time-integrals and split the total mechanical energy loss algebraically.

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