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Question #43092

Difficulty: 8/10 | 🟢 Free

The Question

A plane electromagnetic wave of wavelength $\lambda$ and electric field amplitude $E_0$ is incident from vacuum ( $z < 0$ ) onto a perfectly conducting flat surface at $z=0$ . The wave vector $\vec{k}_i$ lies in the $xz$ -plane and makes an angle $\theta$ ( $0 < \theta < \pi/2$ ) with the $z$ -axis. The electric field is polarized parallel to the $y$ -axis. A small circular wire loop of area $A$ and negligible self-inductance, having resistance $R$ , is placed such that its center is at a distance $d$ from the conducting surface ( $z = -d$ ). The loop lies entirely in a plane parallel to the $xy$ -plane. Assume the dimensions of the loop are much smaller than $\lambda$ . Let $\omega$ be the angular frequency of the wave and $c$ be the speed of light. Determine the maximum possible magnitude of the time-averaged torque $\langle \tau \rangle_{max}$ acting on the loop as $d$ is varied, and the minimum positive distance $d_{max}$ at which this maximum torque occurs.

Options

A.
$\langle \tau \rangle_{max} = \frac{E_0^2 A^2 \omega}{2 R c^2} \sin(2\theta)$ at $d_{max} = \frac{\lambda}{8\cos\theta}$
B.
$\langle \tau \rangle_{max} = \frac{E_0^2 A^2 \omega}{R c^2} \sin(2\theta)$ at $d_{max} = \frac{\lambda}{4\cos\theta}$
C.
$\langle \tau \rangle_{max} = \frac{E_0^2 A^2 \omega}{2 R c^2} \sin(2\theta)$ at $d_{max} = \frac{\lambda}{4\cos\theta}$
D.
$\langle \tau \rangle_{max} = \frac{E_0^2 A^2 \omega}{R c^2} \sin^2\theta$ at $d_{max} = \frac{\lambda}{8\cos\theta}$

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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

Transverse Nature of EM Waves|Magnetic Dipole Moment Orientation|Boundary Condition Reflection

Why This Question Is Tricky

Common Mistakes

This question tests the understanding of electromagnetic standing waves and the vector nature of magnetic torque.

A common trap is assuming that the maximum torque occurs at the position where the induced magnetic dipole moment is maximized.

The dipole moment is driven by the perpendicular magnetic field component ( $B_z$ ), which has antinodes at $d = \lambda / (4\cos\theta)$ .

However, the torque ( $\vec{\tau} = \vec{m} \times \vec{B}$ ) also requires the transverse magnetic field component ( $B_x$ ).

Due to the boundary conditions at the perfect conductor, $B_x$ and $B_z$ are spatially out of phase by $\pi/2$ ; where $B_z$ has an antinode, $B_x$ has a node.

Consequently, the torque is exactly zero at the position of maximum dipole moment.

The true maximum torque occurs halfway between the node and antinode, at $d = \lambda / (8\cos\theta)$ , maximizing the product $\sin(k_z d)\cos(k_z d)$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Waves - Class 12
Concept	Transverse Nature of EM Waves
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free