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Question #43934

Difficulty: 8/10 | 🟢 Free

The Question

A conducting rod of mass mm and length ll rests on two long, parallel, horizontal conducting rails separated by a distance ll. A uniform vertical magnetic field BB exists throughout the region. The rails are connected in series to a DC voltage source V0V_0, a resistor RR, an inductor LL, and a switch. The coefficients of static and kinetic friction between the rod and the rails are μs\mu_s and μk\mu_k, respectively (μs>μk\mu_s > \mu_k). Additionally, when moving, the rod experiences an aerodynamic drag force Fd=−kvF_d = -kv, where vv is its velocity and kk is a positive constant. The switch is closed at t=0t=0. Assuming V0lB>μsmgRV_0 l B > \mu_s mg R, the rod eventually starts moving. Let a0a_0 be the acceleration of the rod immediately after it begins to move, and vfv_f be its steady-state terminal velocity. Which of the following expressions correctly gives the pair (a0,vf)(a_0, v_f)?

Options

  • A.

    ((μs−μk)g,V0lB−μkmgRB2l2+kR)\left( (\mu_s - \mu_k)g, \frac{V_0 l B - \mu_k mg R}{B^2 l^2 + k R} \right)

  • B.

    (V0lB−μkmgRmR,V0lB−μkmgRB2l2+kR)\left( \frac{V_0 l B - \mu_k mg R}{m R}, \frac{V_0 l B - \mu_k mg R}{B^2 l^2 + k R} \right)

  • C.

    ((μs−μk)g,V0lB−μkmgRkR)\left( (\mu_s - \mu_k)g, \frac{V_0 l B - \mu_k mg R}{k R} \right)

  • D.

    (V0lB−μsmgRmR,V0lB−μsmgRB2l2+kR)\left( \frac{V_0 l B - \mu_s mg R}{m R}, \frac{V_0 l B - \mu_s mg R}{B^2 l^2 + k R} \right)

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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

Transient Current in RL Circuit|Kinetic Friction Transition|Differential Equations

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of mechanical and electromagnetic transients, specifically targeting the misapplication of initial conditions and steady-state limits.

A common trap is ignoring the inductor's property of maintaining current continuity at the exact moment static friction is broken, leading students to incorrectly use the asymptotic current V0/RV_0/R for the initial acceleration (Option B).

Another trap is neglecting the motional EMF (Faraday's Law) in the steady-state force balance, which incorrectly decouples the mechanical drag from the electromagnetic damping (Option C).

The correct approach requires recognizing that the inductor 'locks in' the threshold current during the friction transition, while the motional EMF fundamentally alters the steady-state current.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptTransient Current in RL Circuit
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree