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Question #43402

Difficulty: 8/10 | 🟢 Free

The Question

A rigid circular wire loop of radius RR and cross-sectional area AA lies in the x−yx-y plane, centered at the origin. The wire contains a free electron density nn, and the electrons have a relaxation time τ\tau and mass mem_e. A uniform azimuthal electric field E0E_0 is established in the wire, driving a steady base conventional current I0I_0 in the counter-clockwise direction (+θ^+\hat{\theta}). The entire loop is placed inside a rocket that is accelerating with a constant acceleration a⃗=a0i^\vec{a} = a_0 \hat{i} relative to an inertial frame. Assuming steady state is reached in the rocket's frame, find the net magnetic field B⃗\vec{B} at a point (0,0,z)(0, 0, z) on the axis of the loop. (Given: ee is the magnitude of the elementary charge, μ0\mu_0 is the vacuum permeability. Ignore the self-magnetic field's effect on the drift velocity).

Options

  • A.

    μ04(R2+z2)3/2[−nAeτa0zRj^+2I0R2k^]\frac{\mu_0}{4(R^2+z^2)^{3/2}} \left[ -n A e \tau a_0 z R \hat{j} + 2 I_0 R^2 \hat{k} \right]

  • B.

    μ04(R2+z2)3/2[nAeτa0zRj^+2I0R2k^]\frac{\mu_0}{4(R^2+z^2)^{3/2}} \left[ n A e \tau a_0 z R \hat{j} + 2 I_0 R^2 \hat{k} \right]

  • C.

    μ04(R2+z2)3/2[−nAeτa0zRi^+2I0R2k^]\frac{\mu_0}{4(R^2+z^2)^{3/2}} \left[ -n A e \tau a_0 z R \hat{i} + 2 I_0 R^2 \hat{k} \right]

  • D.

    μ04(R2+z2)3/2[2I0R2k^]\frac{\mu_0}{4(R^2+z^2)^{3/2}} \left[ 2 I_0 R^2 \hat{k} \right]

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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

Magnetic Field on Axis vs. Center|Pseudo Force Directionality|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of non-inertial mechanics with electrodynamics via the Drude model.

Students commonly fall into the trap of treating current as a rigid, scalar property (Option D), ignoring that charge carriers possess mass and are subject to pseudo-forces in an accelerating frame.

Even if the pseudo-force is recognized, a secondary trap lies in the sign convention: failing to account for the electron's negative charge when converting drift velocity to conventional current leads to an inverted transverse field (Option B).

Finally, the problem forces the abandonment of the standard axial Biot-Savart formula, requiring a full vector path integral to reveal the symmetry-breaking transverse magnetic field component (j^\hat{j}) generated by the acceleration.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptMagnetic Field on Axis vs. Center
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree