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Question #42085

Difficulty: 8/10 | 🟢 Free

The Question

A small bead of mass mm (in kg\text{kg}) and charge q>0q > 0 (in C\text{C}) is threaded on a long, straight, frictionless horizontal wire lying along the xx-axis. The entire system is immersed in a region containing a time-varying electric field E⃗=E0i^+αt3j^\vec{E} = E_0 \hat{i} + \alpha t^3 \hat{j} (in V⋅m−1\text{V}\cdot\text{m}^{-1}) and a non-uniform magnetic field B⃗=B0xk^\vec{B} = B_0 x \hat{k} (in T\text{T}), where E0E_0 and B0B_0 are positive constants, α\alpha is an unknown constant, and tt is time (in s\text{s}). The bead is released from rest at the origin at t=0t=0. If the normal force exerted by the wire on the bead is zero at all times, what is the value of the constant α\alpha?

Options

  • A.

    q2E02B02m2\frac{q^2 E_0^2 B_0}{2 m^2}

  • B.

    q2E02B0m2\frac{q^2 E_0^2 B_0}{m^2}

  • C.

    −q2E02B02m2-\frac{q^2 E_0^2 B_0}{2 m^2}

  • D.

    qE0B02m\frac{q E_0 B_0}{2 m}

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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

Velocity Vector Direction|Lorentz Force Vector Direction|Integration

Why This Question Is Tricky

Common Mistakes

This problem tests the non-trivial coupling of a time-varying electric field and a spatially non-uniform magnetic field under a kinematic constraint.

A common trap is assuming the magnetic field alters the longitudinal acceleration or kinetic energy, whereas it strictly acts perpendicular to the instantaneous velocity vector.

Students must correctly integrate the constant longitudinal acceleration to find both instantaneous velocity and position, and then apply the Lorentz force cross-product (specifically noting that i^×k^=−j^\hat{i} \times \hat{k} = -\hat{j}) to balance the time-varying transverse electric force.

Mistakes in the cross-product sign lead to a negative α\alpha, while incorrectly using x=vtx=vt instead of x=12at2x=\frac{1}{2}at^2 misses the factor of 2 in the denominator.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Plane - Class 11 Part 1
ConceptVelocity Vector Direction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree