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Question #42430

Difficulty: 8/10 | 🟢 Free

The Question

A small square loop of mass mm, side length aa, and electrical resistance RR is constrained to move along the positive xx-axis on a smooth horizontal surface. A light inextensible string is attached to the loop, passes over a fixed massless pulley at x=Dx = D, then goes around a movable massless pulley, and is finally attached to a fixed peg at x=Dx = D. All string segments are parallel to the xx-axis. A block of mass MM is attached to the movable pulley and slides on the same horizontal surface. A non-uniform magnetic field B⃗=B0x2k^\vec{B} = B_0 x^2 \hat{k} exists in the region. At t=0t=0, the loop is at x=0x=0 and is given an initial velocity V0V_0 in the +x+x direction. Find the maximum xx-coordinate reached by the loop before it momentarily comes to rest. (Assume a≪xa \ll x and DD is sufficiently large)

Options

  • A.

    (3R(4m+M)V016B02a4)1/3\left( \frac{3 R (4m + M) V_0}{16 B_0^2 a^4} \right)^{1/3}

  • B.

    (3R(m+M)V04B02a4)1/3\left( \frac{3 R (m + M) V_0}{4 B_0^2 a^4} \right)^{1/3}

  • C.

    (3R(m+4M)V04B02a4)1/3\left( \frac{3 R (m + 4M) V_0}{4 B_0^2 a^4} \right)^{1/3}

  • D.

    (3R(4m+M)V08B02a4)1/3\left( \frac{3 R (4m + M) V_0}{8 B_0^2 a^4} \right)^{1/3}

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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

Constraint Relations via String Length|Magnetic Dipole Moment Orientation|Calculus/Integration

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of kinematic constraints with electrodynamic damping in a non-uniform magnetic field.

A common trap is incorrectly applying the string constraint to the equivalent inertia of the system (yielding m+Mm+M or m+4Mm+4M instead of m+M/4m+M/4).

Furthermore, students often miscalculate the position-dependent magnetic force by either using the absolute field BB instead of the gradient dB/dxdB/dx across the loop, or by failing to properly integrate the non-linear velocity-dependent deceleration using the work-energy theorem or the chain rule dv/dt=v(dv/dx)dv/dt = v(dv/dx).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterLaws of Motion - Class 11 Part 1
ConceptConstraint Relations via String Length
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree