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Question #42673

Difficulty: 8/10 | 🟢 Free

The Question

A long cylindrical conducting shell of inner radius R1R_1 and outer radius R2R_2 is placed in a uniform magnetic field B⃗\vec{B} pointing in the +z+z direction. The magnetic field exists everywhere in space and its magnitude varies with time as B(t)=αtB(t) = \alpha t, where α\alpha is a positive constant. A point charge qq is moved slowly by an external agent from point PP to point QQ through the conducting material. In cylindrical coordinates (r,ϕ,z)(r, \phi, z) with the z-axis along the cylinder's axis, the coordinates of PP and QQ are (R2,0,0)(R_2, 0, 0) and (R2,π,0)(R_2, \pi, 0) respectively.

Consider two paths for moving the charge: Path A: A semi-circular arc along the outer surface of the shell (r=R2r = R_2). Path B: A path consisting of three segments: radially inward from PP to (R1,0,0)(R_1, 0, 0), then a semi-circular arc along the inner surface (r=R1r = R_1) to (R1,π,0)(R_1, \pi, 0), and finally radially outward to QQ.

Assuming the charge is moved slowly enough that radiation and self-magnetic field effects are negligible, what is the difference in the work done by the external agent along Path A and Path B, i.e., ΔW=WA−WB\Delta W = W_A - W_B?

Options

  • A.

    πqα2(R22−R12)\frac{\pi q \alpha}{2} (R_2^2 - R_1^2)

  • B.

    00

  • C.

    πqα(R22−R12)\pi q \alpha (R_2^2 - R_1^2)

  • D.

    πqα2(R22+R12)\frac{\pi q \alpha}{2} (R_2^2 + R_1^2)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Conductor Interior Potential|Eddy Currents Dissipation|Conservative vs Non-conservative

Why This Question Is Tricky

Common Mistakes

This question targets the misconception that the electric field inside a conductor is always zero, leading students to incorrectly assume that the work done to move a charge through a conducting medium is zero (or path-independent).

While true in electrostatics, a time-varying magnetic field induces a non-conservative electric field (∇×E⃗≠0\nabla \times \vec{E} \neq 0).

Because the conductor has finite resistivity, this induced field drives eddy currents (J⃗=σE⃗\vec{J} = \sigma \vec{E}) and is not canceled out by surface charges.

Consequently, the work done by an external agent depends on the path taken, violating the concept of a scalar electric potential.

The trap option 0 catches students who blindly apply ΔV=0\Delta V = 0 inside conductors without accounting for Faraday's Law of Induction.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptConductor Interior Potential
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree