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Question #43599

Difficulty: 8/10 | 🟢 Free

The Question

A rigid square conducting loop of side $L$ and total resistance $R$ lies in the $xy$ -plane. At $t=0$ , it is pulled with a constant velocity $v_0 \hat{i}$ into a region $x > 0$ containing a uniform magnetic field $B_0 \hat{k}$ . Mounted on the loop is an optical system consisting of a fixed polarizer and a second polarizer rotating at a constant angular velocity $\omega$ . Unpolarized light of intensity $2I_0$ is incident on the system, such that the transmitted intensity is $I(t) = I_0 \cos^2(\omega t)$ . A transducer on the loop uses this light to generate an additional uniform magnetic field $B_{int}(t) = \beta I(t) \hat{k}$ strictly confined within the entire area $L^2$ of the loop. Neglecting self-inductance, what is the average electrical power dissipated in the loop as heat during the time it is partially inside the external magnetic field? (Assume the time taken to enter is much larger than $2\pi/\omega$ ).

Options

A.
$\frac{B_0^2 L^2 v_0^2}{R}$
B.
$\frac{B_0^2 L^2 v_0^2}{R} + \frac{\beta^2 I_0^2 L^4 \omega^2}{2R}$
C.
$\frac{B_0^2 L^2 v_0^2}{R} + \frac{\beta^2 I_0^2 L^4 \omega^2}{R}$
D.
$\frac{B_0^2 L^2 v_0^2}{R} + \frac{\beta^2 I_0^2 L^4 \omega^2}{4R}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Malus's Law|Lenz's Law Energy Conservation|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests a lethal misconception regarding the Work-Energy Theorem in electrodynamics.

Students often blindly assume that the average electrical power dissipated in a moving loop equals the average mechanical power supplied by the external pulling force ( $\langle P_{mech} \rangle = \langle F_{ext} v_0 \rangle$ ).

If they calculate the magnetic drag force $F_{mag} = i L B_0$ , they will find $\langle P_{mech} \rangle = \frac{B_0^2 L^2 v_0^2}{R}$ , leading them straight to the trap Option A.

They fail to recognize that the transducer acts as an active AC power source driven by the light energy.

Because the transformer EMF fluctuates symmetrically, it contributes zero net average force against the constant velocity, but its RMS value contributes significantly to the $i^2 R$ Joule heating.

The problem demands a rigorous, first-principles application of Faraday's Law and Joule's Law rather than a misapplied global energy conservation shortcut.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Malus's Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free