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Question #42970

Difficulty: 8/10 | 🟢 Free

The Question

A conducting rod of mass mm (in kg) and length LL (in m) slides down a rough inclined plane of angle θ\theta with the horizontal. The plane has a coefficient of kinetic friction μ\mu. The rod is in contact with two parallel, perfectly conducting rails fixed to the incline, which are connected at the bottom by a resistor of resistance RR (in Ω\Omega). A uniform magnetic field of magnetic induction BB (in T) exists vertically downwards throughout the region. The rod is released from rest. Assuming the rails are very long, what is the terminal velocity v0v_0 (in m/s) acquired by the rod? (Neglect the self-inductance of the loop and electrical resistance of the rod and rails).

Options

  • A.

    mgR(sin⁡θ−μcos⁡θ)B2L2cos⁡θ(cos⁡θ+μsin⁡θ)\frac{mg R (\sin\theta - \mu \cos\theta)}{B^2 L^2 \cos\theta (\cos\theta + \mu \sin\theta)}

  • B.

    mgR(sin⁡θ−μcos⁡θ)B2L2cos⁡2θ\frac{mg R (\sin\theta - \mu \cos\theta)}{B^2 L^2 \cos^2\theta}

  • C.

    mgR(sin⁡θ−μcos⁡θ)B2L2\frac{mg R (\sin\theta - \mu \cos\theta)}{B^2 L^2}

  • D.

    mgR(sin⁡θ−μcos⁡θ)B2L2cos⁡θ(cos⁡θ−μsin⁡θ)\frac{mg R (\sin\theta - \mu \cos\theta)}{B^2 L^2 \cos\theta (\cos\theta - \mu \sin\theta)}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Magnetic Flux Scalar Sign|Eddy Currents Dissipation|Wedge Constraint Logic

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of electromagnetic induction with non-conservative mechanical forces via a modified normal force constraint.

The primary trap is the assumption that the normal force on an incline remains mgcos⁡θmg \cos\theta.

Students often correctly identify the horizontal magnetic braking force FmF_m but fail to resolve its component perpendicular to the incline.

Because FmF_m is horizontal and opposes the horizontal component of velocity, it pushes the rod into the wedge, increasing the normal force to N=mgcos⁡θ+Fmsin⁡θN = mg \cos\theta + F_m \sin\theta.

This creates a secondary, velocity-dependent braking effect through increased kinetic friction, fundamentally altering the terminal velocity equation.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptMagnetic Flux Scalar Sign
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree