Before You Begin

Question #42476

Difficulty: 8/10 | 🟢 Free

The Question

Two parallel, horizontal, frictionless conducting rails are separated by a distance $L$ and connected at $x=0$ by a resistor of resistance $R$ . A conducting rod of mass $m$ and length $L$ is placed across the rails. A spatially uniform magnetic field $\vec{B}$ is perpendicular to the plane of the rails. The magnitude of the magnetic field varies with time $t$ as $B(t) = B_0 (1 - \alpha t)$ , where $B_0$ and $\alpha$ are positive constants. An external agent pulls the rod such that it moves with a constant velocity $v_0$ in the $+x$ direction. At $t=0$ , the rod is at $x=0$ . Assuming the self-inductance of the loop is negligible, what is the total work done by the external agent from $t=0$ to $t = 1/\alpha$ ?

Options

A.
$\frac{B_0^2 L^2 v_0^2}{6 \alpha R}$
B.
$\frac{B_0^2 L^2 v_0^2}{3 \alpha R}$
C.
$\frac{B_0^2 L^2 v_0^2}{2 \alpha R}$
D.
$\frac{B_0^2 L^2 v_0^2}{4 \alpha R}$

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

Work-Energy Theorem Validity|Lenz’s Law Energy Conservation|Motional EMF for Arbitrary Shapes

Why This Question Is Tricky

Common Mistakes

This question targets a profound misconception in electromagnetic induction: the assumption that the work done by external mechanical forces always equals the Joule heating in a moving conductor ( $W_{ext} = H$ ).

While true for static magnetic fields, this equality fails when the magnetic field is time-varying.

A time-varying magnetic field generates a non-conservative induced electric field ( $\nabla \times \vec{E} \neq 0$ ) throughout space.

This field performs direct electrical work on the charge carriers in both the moving rod and the stationary rails.

Consequently, the generalized Work-Energy Theorem must include this non-mechanical work: $W_{ext} + W_{nc} = H$ .

Students who blindly equate mechanical work to heat dissipation will incorrectly calculate $H$ and choose the trap option, completely missing the field-mediated energy transfer.

Quick Facts About This Question

Question Information

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