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Question #43665

Difficulty: 8/10 | 🟢 Free

The Question

Two small particles, P1P_1 and P2P_2, having masses mm and 2m2m, and charges +q+q and +2q+2q respectively, are connected by an ideal massless spring of spring constant kk and natural length zero. The system is placed in a uniform magnetic field B⃗=Bk^\vec{B} = B \hat{k}. The particles are projected such that they undergo steady circular motion in the xyxy-plane about their common center of mass, which remains at rest. If the particles rotate in the counter-clockwise direction, what is the steady angular frequency ω\omega of their relative motion?

Options

  • A.

    q2B2+6mk−qB2m\frac{\sqrt{q^2 B^2 + 6mk} - qB}{2m}

  • B.

    q2B2+6mk+qB2m\frac{\sqrt{q^2 B^2 + 6mk} + qB}{2m}

  • C.

    q2B2+4mk−qB2m\frac{\sqrt{q^2 B^2 + 4mk} - qB}{2m}

  • D.

    9q2B2+24mk−3qB4m\frac{\sqrt{9q^2 B^2 + 24mk} - 3qB}{4m}

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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

Reduced Mass Effect|Trajectory of Charged Particle|Conservative Field Path Independence|Vector Analysis

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Lorentz force in a two-body system.

A common trap is to incorrectly apply the magnetic force to the relative motion by either using the rest mass mm instead of the reduced mass μ\mu, or by failing to derive the effective charge qeff=μ(q/m)q_{eff} = \mu (q/m) for the relative frame.

Furthermore, students often miscalculate the direction of the cross product v⃗rel×B⃗\vec{v}_{rel} \times \vec{B}, failing to recognize that it generates an outward 'centrifugal-like' force for counter-clockwise rotation, leading to a sign error in the radial differential equation.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAtoms - Class 12
ConceptReduced Mass Effect
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree