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Question #42973

Difficulty: 8/10 | 🟢 Free

The Question

Two long, parallel conducting rails of zero resistance, separated by a distance LL, are fixed on an inclined plane at an angle θ\theta to the horizontal. The rails are connected at the bottom by a conducting wire of zero resistance. A uniform magnetic field B⃗\vec{B} points vertically downwards. A conducting rod of mass mm and resistance RR is placed horizontally across the rails. The coefficient of kinetic friction between the rod and the rails is μ\mu. A constant external force FF pulls the rod up the incline, parallel to the rails. The rod starts from rest. Which of the following expressions gives the distance xx moved by the rod along the incline when it attains a velocity vv? (Assume the rails are long enough and FF is sufficiently large to move the rod upwards. Let Feff=F−mg(sin⁡θ+μcos⁡θ)F_{eff} = F - mg(\sin\theta + \mu \cos\theta)).

Options

  • A.

    x=mFeffβ2ln⁡(FeffFeff−βv)−mvβx = \frac{m F_{eff}}{\beta^2} \ln\left(\frac{F_{eff}}{F_{eff} - \beta v}\right) - \frac{m v}{\beta}, where β=B2L2cos⁡θR(cos⁡θ+μsin⁡θ)\beta = \frac{B^2 L^2 \cos\theta}{R} (\cos\theta + \mu \sin\theta)

  • B.

    x=mFeffβ2ln⁡(FeffFeff−βv)−mvβx = \frac{m F_{eff}}{\beta^2} \ln\left(\frac{F_{eff}}{F_{eff} - \beta v}\right) - \frac{m v}{\beta}, where β=B2L2cos⁡2θR\beta = \frac{B^2 L^2 \cos^2\theta}{R}

  • C.

    x=mFeffβ2ln⁡(FeffFeff−βv)−mvβx = \frac{m F_{eff}}{\beta^2} \ln\left(\frac{F_{eff}}{F_{eff} - \beta v}\right) - \frac{m v}{\beta}, where β=B2L2R\beta = \frac{B^2 L^2}{R}

  • D.

    x=mFeffβ2ln⁡(FeffFeff−βv)−mvβx = \frac{m F_{eff}}{\beta^2} \ln\left(\frac{F_{eff}}{F_{eff} - \beta v}\right) - \frac{m v}{\beta}, where β=B2L2cos⁡θR(cos⁡θ−μsin⁡θ)\beta = \frac{B^2 L^2 \cos\theta}{R} (\cos\theta - \mu \sin\theta)

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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

Lenz’s Law Energy Conservation|Work Done by Friction|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple non-conservative frictional forces with velocity-dependent electromagnetic forces.

The primary trap lies in the formulation of the normal force.

Students often mechanically write the normal force on an incline as N=mgcos⁡θN = mg \cos\theta, forgetting that the Lorentz force F⃗m=IL⃗×B⃗\vec{F}_m = I \vec{L} \times \vec{B} is strictly horizontal because the magnetic field is vertical.

This horizontal magnetic force has a component Fmsin⁡θF_m \sin\theta perpendicular to the incline, which presses the rod harder against the rails, dynamically increasing the frictional force as velocity increases.

Failing to account for this feedback loop leads to an incorrect damping coefficient β\beta (Option B or D), testing the synthesis of rigorous rigid-body mechanics with electromagnetism.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptLenz’s Law Energy Conservation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree