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Question #42532

Difficulty: 8/10 | 🟢 Free

The Question

A thin circular conducting plate of radius $R$ , thickness $h$ , and electrical conductivity $\sigma_c$ is placed in a uniform magnetic field perpendicular to its plane. The magnetic field varies with time $t$ as $B(t) = \alpha t$ , where $\alpha$ is a positive constant. The plate is in a vacuum and its radial thermal conductivity is negligible. The plate radiates heat from both its flat surfaces. The surfaces are non-Lambertian, with a directional emissivity given by $\epsilon(\theta) = \epsilon_0 \cos\theta$ , where $\theta$ is the angle from the surface normal and $\epsilon_0$ is a constant. If $\sigma$ is the Stefan-Boltzmann constant, what is the steady-state temperature $T(R)$ at the rim of the plate?

Options

A.
$\left( \frac{3 \sigma_c h \alpha^2 R^2}{16 \epsilon_0 \sigma} \right)^{\frac{1}{4}}$
B.
$\left( \frac{3 \sigma_c h \alpha^2 R^2}{8 \epsilon_0 \sigma} \right)^{\frac{1}{4}}$
C.
$\left( \frac{\sigma_c h \alpha^2 R^2}{8 \epsilon_0 \sigma} \right)^{\frac{1}{4}}$
D.
$\left( \frac{3 \sigma_c h \alpha^2 R^2}{32 \epsilon_0 \sigma} \right)^{\frac{1}{4}}$

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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

Emissivity and Absorptivity|Lenz’s Law Energy Conservation|Vector Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple electromagnetic induction with advanced radiative heat transfer.

The primary trap is assuming a uniform temperature distribution (lumped capacitance model) and equating total power instead of applying a local power balance, which is required because the eddy-current heating is radially dependent (proportional to r^2) and radial conduction is negligible.

A secondary trap is the misapplication of the Stefan-Boltzmann law for a non-Lambertian surface; students often incorrectly use the constant directly instead of integrating the directional emissivity over the hemisphere to find the effective hemispherical emissivity.

Finally, forgetting that the plate radiates from both surfaces introduces a factor of 2 error.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermal Properties of Matter - Class 11 Part 2
Concept	Emissivity and Absorptivity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free