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Question #42763

Difficulty: 8/10 | 🟢 Free

The Question

A thin, uncharged conducting spherical shell of radius $R$ , centered at the origin, is placed in a uniform time-varying magnetic field $\vec{B} = \alpha t \hat{k}$ , where $\alpha$ is a positive constant and $t$ is time. Neglecting the magnetic field produced by the induced eddy currents, what is the electric field $\vec{E}$ at a point $P$ inside the cavity having position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ ? (Assume $x^2+y^2+z^2 < R^2$ )

Options

A.
$\vec{0}$
B.
$\frac{\alpha}{2} (y\hat{i} - x\hat{j})$
C.
$-\frac{\alpha}{2} (x\hat{i} + y\hat{j})$
D.
$\alpha (y\hat{i} - x\hat{j})$

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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

Cavity Field Shielding|Induced Electric Fields|Faraday’s Law Vector Form

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental limits of electrostatic shielding, a concept frequently misapplied by students.

A common misconception is that a conducting cavity is always field-free ( $\vec{E} = \vec{0}$ ) under all circumstances.

However, electrostatic shielding only applies to conservative electric fields generated by external static charges, where surface charges on the conductor can redistribute to perfectly cancel the internal field (Gauss's Law).

A time-varying magnetic field induces a non-conservative (curl) electric field.

Because this specific induced field is purely tangential to the spherical shell, it causes no charge accumulation.

Consequently, the conductor cannot produce a counter-field, and the induced electric field penetrates the cavity completely.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Cavity Field Shielding
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free