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Question #43876

Difficulty: 8/10 | 🟢 Free

The Question

A pair of long, parallel, perfectly conducting rails are fixed on a horizontal rough plane (the xyxy-plane), separated by a distance LL. The rails are parallel to the yy-axis and are connected at y=0y=0 by a wire of zero resistance. A uniform metal rod of mass mm, length LL, and resistance RR is placed perpendicularly across the rails. The coefficient of kinetic friction between the rod and the rails is μ\mu. A uniform magnetic field B⃗=B0j^−B0k^\vec{B} = B_0 \hat{j} - B_0 \hat{k} exists in the entire region, where j^\hat{j} is the unit vector along the positive yy-axis and k^\hat{k} is the unit vector vertically upwards. The acceleration due to gravity is gg in the −k^-\hat{k} direction. At t=0t=0, the rod is located at y>0y > 0 and is given an initial velocity v0j^v_0 \hat{j}. Assuming the rod remains in contact with the rails at all times, which of the following expressions gives the time TT taken by the rod to come to rest?

Options

  • A.

    mRB02L2(1+μ)ln⁡(1+B02L2(1+μ)v0μmgR)\frac{m R}{B_0^2 L^2 (1 + \mu)} \ln \left( 1 + \frac{B_0^2 L^2 (1 + \mu) v_0}{\mu m g R} \right)

  • B.

    mRB02L2ln⁡(1+B02L2v0μmgR)\frac{m R}{B_0^2 L^2} \ln \left( 1 + \frac{B_0^2 L^2 v_0}{\mu m g R} \right)

  • C.

    mRB02L2(1−μ)ln⁡(1+B02L2(1−μ)v0μmgR)\frac{m R}{B_0^2 L^2 (1 - \mu)} \ln \left( 1 + \frac{B_0^2 L^2 (1 - \mu) v_0}{\mu m g R} \right)

  • D.

    mR(1+μ)B02L2ln⁡(1+B02L2v0μmgR(1+μ))\frac{m R (1 + \mu)}{B_0^2 L^2} \ln \left( 1 + \frac{B_0^2 L^2 v_0}{\mu m g R (1 + \mu)} \right)

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Force on Current-Carrying Loop|Frame of Reference Dependency|Kinetic Friction Transition

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize that the normal force is not always simply mgmg, especially in electromagnetism problems involving multi-dimensional magnetic fields.

The primary trap is the assumption that the magnetic force only acts to oppose the motion (horizontal drag).

Because the magnetic field has a component parallel to the velocity (j^\hat{j}), the induced current (which is along −i^-\hat{i}) interacts with this parallel component to produce a velocity-dependent vertical force.

This dynamic vertical force continuously alters the normal force and, consequently, the kinetic friction.

Students who blindly apply fk=μmgf_k = \mu mg will select Option B, completely missing the coupled electro-mechanical nature of the friction in this specific frame.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptForce on Current-Carrying Loop
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree