Start Learning Free

Before You Begin

Question #43844

Difficulty: 8/10 | 🟢 Free

The Question

A thin n-type semiconductor ring of radius RR, radial width ww, and axial thickness tt (w,t≪Rw, t \ll R) lies in the xyxy-plane, centered at the origin. A magnetic dipole of moment m⃗=mi^\vec{m} = m \hat{i} is fixed at the origin. A total steady current II is driven around the ring in the +ϕ^+\hat{\phi} direction. The charge carriers have mobility μ\mu and charge magnitude ee.

Due to the spatially varying magnetic field, a Hall electric field EzE_z develops in the axial direction. To satisfy steady-state boundary conditions (∇×E⃗=0\nabla \times \vec{E} = 0), the azimuthal current density JϕJ_\phi must vary across the thickness of the ring.

Assuming the variation in drift velocity is small compared to its average value, find the magnitude of the difference in azimuthal current density, ΔJ=∣Jϕ(z=t/2)−Jϕ(z=−t/2)∣\Delta J = |J_\phi(z=t/2) - J_\phi(z=-t/2)|, at the angular position ϕ=π/2\phi = \pi/2 (on the yy-axis).

Options

  • A.

    μμ0mI2πwR4\frac{\mu \mu_0 m I}{2\pi w R^4}

  • B.

    μμ0mIπwR4\frac{\mu \mu_0 m I}{\pi w R^4}

  • C.

    μμ0mI4πwR4\frac{\mu \mu_0 m I}{4\pi w R^4}

  • D.

    0

Check Your Answer

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

Drift vs Diffusion Current|Trajectory of Charged Particle|Magnetic Field on Axial vs Equatorial Line

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of the Hall effect, Faraday's law of induction (∇×E⃗=0\nabla \times \vec{E} = 0 in steady state), and the continuity equation.

Students typically memorize that steady currents in uniform wires have uniform current density and might incorrectly choose 0.

The trap lies in the inhomogeneous magnetic field of the dipole, which creates a spatially varying transverse Hall field.

To prevent a non-zero curl of the electric field, the longitudinal driving field (and thus the current density) must develop a transverse gradient.

Failing to couple the orthogonal electric field components via ∇×E⃗=0\nabla \times \vec{E} = 0 leads to a fundamental misunderstanding of steady-state electrodynamics in non-uniform fields.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptDrift vs Diffusion Current
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree