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Question #42713

Difficulty: 8/10 | 🟢 Free

The Question

A uniform conducting wire of constant cross-sectional area AA and free electron density nn is bent into a semi-circle of radius RR. A time-varying voltage is applied across its ends such that the tangential electric field inside the wire is E(t)=E0sin⁡(ωt)E(t) = E_0 \sin(\omega t) for 0≤t≤π/ω0 \leq t \leq \pi/\omega, where E0E_0 is a positive constant. An average conduction electron enters one end of the semi-circle at t=0t=0 and exactly reaches the other end at t=π/ωt = \pi/\omega. Let μ\mu be the electron mobility and ee be the elementary charge. Which of the following gives the magnitude of the average drift velocity vector ∣⟨v⃗d⟩∣|\langle \vec{v}_d \rangle| of this electron during this time interval, and the total charge QQ that flows through any cross-section of the wire?

Options

  • A.

    ∣⟨v⃗d⟩∣=4μE0π2,  Q=πRneA|\langle \vec{v}_d \rangle| = \frac{4\mu E_0}{\pi^2}, \; Q = \pi R n e A

  • B.

    ∣⟨v⃗d⟩∣=2μE0π,  Q=πRneA|\langle \vec{v}_d \rangle| = \frac{2\mu E_0}{\pi}, \; Q = \pi R n e A

  • C.

    ∣⟨v⃗d⟩∣=4μE0π2,  Q=2RneA|\langle \vec{v}_d \rangle| = \frac{4\mu E_0}{\pi^2}, \; Q = 2 R n e A

  • D.

    ∣⟨v⃗d⟩∣=2μE0π,  Q=2RneA|\langle \vec{v}_d \rangle| = \frac{2\mu E_0}{\pi}, \; Q = 2 R n e A

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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

Drift Velocity Mechanism|Displacement vs. Total Path Length|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between microscopic path length and macroscopic displacement in the context of drift kinematics and the continuity equation.

Students often erroneously calculate the average drift velocity vector using the total path length along the curved conductor (yielding 2μE0/π2\mu E_0/\pi, which is actually the average speed) instead of the net straight-line displacement (2R2R).

Conversely, when calculating the total charge flow, students might incorrectly use the displacement instead of the path length.

The correct physical model requires using displacement for the average velocity vector, but the path length (arc length) for the charge flux, as current is constrained to flow along the wire's geometry.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptDrift Velocity Mechanism
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree