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Question #43398

Difficulty: 8/10 | 🟢 Free

The Question

A U-shaped conducting frame with parallel rails separated by a distance LL is fixed on an inclined plane of angle θ\theta. The rails are connected at the top by a resistor of resistance R0R_0. A conducting rod of mass mm can slide down the rails. The coefficient of kinetic friction between the rod and the rails is μ\mu. Let the incline surface define the x−zx-z plane, with the xx-axis pointing directly down the incline and the zz-axis horizontal. The top edge of the frame lies on the zz-axis (x=0x=0). The yy-axis is perpendicular to the incline, pointing upwards. An infinitely long straight wire carrying a constant current II in the +z+z direction is fixed parallel to the zz-axis at a position (x=0,y=−d)(x=0, y=-d). The rod is pulled down the incline at a constant velocity vv by an external force. Assuming the resistance of the rod and rails is negligible, find the magnitude of the total resistive force (sum of kinetic friction and magnetic drag) acting on the rod when it is at a distance xx down the incline.

Options

  • A.

    μmgcos⁡θ+μ02I2L2xv4π2R0(x2+d2)2(μd+x)\mu mg \cos\theta + \frac{\mu_0^2 I^2 L^2 x v}{4\pi^2 R_0 (x^2 + d^2)^2} (\mu d + x)

  • B.

    μmgcos⁡θ+μ02I2L2x2v4π2R0(x2+d2)2\mu mg \cos\theta + \frac{\mu_0^2 I^2 L^2 x^2 v}{4\pi^2 R_0 (x^2 + d^2)^2}

  • C.

    μmgcos⁡θ+μ02I2L2xv4π2R0(x2+d2)2(d+μx)\mu mg \cos\theta + \frac{\mu_0^2 I^2 L^2 x v}{4\pi^2 R_0 (x^2 + d^2)^2} (d + \mu x)

  • D.

    μmgcos⁡θ+μ02I2L2xv4π2R0(x2+d2)3/2(μd+x)\mu mg \cos\theta + \frac{\mu_0^2 I^2 L^2 x v}{4\pi^2 R_0 (x^2 + d^2)^{3/2}} (\mu d + x)

Check Your Answer

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

Infinite Wire vs. Finite Wire|Friction on Inclined Planes|Vector Calculus

Why This Question Is Tricky

Common Mistakes

Students incorrectly assume that the magnetic force on a sliding rod only produces a drag force opposing the velocity.

They calculate the perpendicular magnetic field component ByB_y to find the induced current and the resulting drag, but completely ignore the parallel magnetic field component BxB_x.

This BxB_x component interacts with the induced current to generate a Lorentz force perpendicular to the incline, which dynamically increases the normal force.

By failing to couple the position-dependent Lorentz force to the normal force, students calculate a constant normal force N=mgcos⁡θN = mg \cos\theta and select the trap option, missing the additional velocity-dependent frictional damping.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptInfinite Wire vs. Finite Wire
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree