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Question #43303

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The Question

A horizontal metallic rod of mass $m$ and length $L$ is released from rest at $t=0$ and falls under gravity (acceleration $g$ ) along two vertical, frictionless, perfectly conducting rails. The rails are connected at the top ( $y=0$ ) by a resistor of resistance $R$ . The entire system is immersed in a uniform, time-varying magnetic field $\vec{B} = B_0 t \hat{k}$ perpendicular to the plane of the rails, where $B_0$ is a positive constant. The rod experiences a viscous drag force of magnitude $F_d = b v$ , where $v$ is its downward speed and $b$ is a positive constant. Let $y$ be the downward displacement of the rod and $I$ be the magnitude of the instantaneous induced current. Which of the following expressions gives the instantaneous rate of loss of mechanical energy of the rod?

Options

A.
$I^2 R + b v^2$
B.
$I^2 R + b v^2 - I L y B_0$
C.
$I^2 R + b v^2 + I L y B_0$
D.
$I^2 R + b v^2 - I L v B_0 t$

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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

Faraday’s Law Vector Form|Mechanical Energy Conservation|Reynolds Number Transition

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Work-Energy Theorem in the presence of time-varying magnetic fields.

A common trap for JEE Advanced students is to blindly assume that the rate of loss of mechanical energy is always equal to the sum of the Joule heating ( $I^2 R$ ) and viscous dissipation ( $bv^2$ ).

However, this is only true for static magnetic fields where the magnetic force does no work and all electrical energy comes from motional EMF.

When the magnetic field is time-varying, the induced non-conservative electric field (transformer EMF) also does work on the charge carriers.

The total Joule heating is driven by both motional ( $vBL$ ) and transformer ( $yL\dot{B}$ ) EMFs.

Therefore, the mechanical energy loss only accounts for the work done against the motional component, requiring the subtraction of the transformer power term ( $I L y B_0$ ) from the total dissipated heat.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	Faraday’s Law Vector Form
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free