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Question #42329

Difficulty: 8/10 | 🟢 Free

The Question

A small horizontal circular metallic loop of radius $r$ , mass $m$ , and resistance $R$ is released from rest at $z=0$ and falls under gravity (acceleration $g$ ) along the positive z-axis (downwards). The magnetic field in the region is axisymmetric and its vertical component is given by $B_z = B_0 (1 + \alpha z)$ , where $B_0$ and $\alpha$ are positive constants. The loop reaches a terminal velocity well before it has fallen a distance $H$ . Assuming the loop remains horizontal and its radius $r$ is small enough that the magnetic field components can be approximated to first order in $r$ , what is the total electrical energy dissipated as Joule heating in the loop during its fall from $z=0$ to $z=H$ ?

Options

A.
$mgH$
B.
$mgH - \frac{m^3 g^2 R^2}{2 \pi^4 r^8 B_0^4 \alpha^4}$
C.
$mgH + \frac{m^3 g^2 R^2}{2 \pi^4 r^8 B_0^4 \alpha^4}$
D.
$\frac{m^3 g^2 R^2}{2 \pi^4 r^8 B_0^4 \alpha^4}$

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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

Mechanical Energy Conservation|Motional EMF for Arbitrary Shapes|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Work-Energy Theorem in the presence of non-conservative velocity-dependent magnetic braking forces.

A common trap for JEE aspirants is assuming that because the loop falls at terminal velocity for the majority of its descent, the total gravitational potential energy lost is entirely dissipated as Joule heating ( $Q = mgH$ ).

This neglects the initial acceleration phase where a portion of the potential energy is converted into the loop's steady-state kinetic energy.

Furthermore, it implicitly tests the understanding that a spatially varying axial magnetic field must have a radial component to satisfy $\nabla \cdot \vec{B} = 0$ , which is the actual source of the vertical magnetic braking force, though it can be elegantly bypassed using power equivalence.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Mechanical Energy Conservation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free