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Question #42981

Difficulty: 8/10 | 🟢 Free

The Question

A rigid conducting wire of arbitrary shape and unknown mass connects the center OO of a stationary vertical circular conducting ring of radius a=2.0 ma = 2.0 \text{ m} to a point on its rim. The ring lies in the vertical xyxy-plane, with uniform gravity acting in the −y-y direction. The center OO and the rim are connected via ideal slip rings to an external stationary resistor R=10 ΩR = 10 \ \Omega. The electrical resistance of the wire and the ring are negligible. A magnetic field B⃗=B0(ra)2k^\vec{B} = B_0 \left( \frac{r}{a} \right)^2 \hat{k} exists in the region, where rr is the radial distance from OO, k^\hat{k} is the unit vector normal to the xyxy-plane, and B0=4.0 TB_0 = 4.0 \text{ T}. An external motor rotates the wire about the zz-axis with a constant angular velocity ω=5.0 rad s−1\omega = 5.0 \text{ rad s}^{-1}. Calculate the total work done by the motor during exactly one complete revolution of the wire.

Options

  • A.

    16π J16\pi \text{ J}

  • B.

    64π J64\pi \text{ J}

  • C.

    8π J8\pi \text{ J}

  • D.

    0 J0 \text{ J}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Motional EMF for Arbitrary Shapes|Conservative vs Non-conservative|Work-Energy Theorem for Rotation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between conservative and non-conservative forces within the Work-Energy Theorem.

A common trap is assuming that because the wire returns to its initial state (constant kinetic energy, zero net change in gravitational potential energy), the net work done by the motor must be zero (0 J0 \text{ J}).

Students fail to recognize that the magnetic braking force is non-conservative and continuously dissipates energy as Joule heating.

Furthermore, the 'arbitrary shape' and 'unknown mass' act as a psychological trap, leading students to believe the problem is intractable via standard Newtonian torque integration (τ⃗=∫r⃗×(Idl⃗×B⃗)\vec{\tau} = \int \vec{r} \times (I d\vec{l} \times \vec{B})).

The elegant path forces a transition from rotational dynamics to electromagnetic power loss, proving that the motional EMF depends only on the radial endpoints.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptMotional EMF for Arbitrary Shapes
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree