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Question #42984

Difficulty: 8/10 | 🟢 Free

The Question

A uniform conducting rod of length L=1.0 mL = 1.0 \text{ m} and resistance R=5.0 ΩR = 5.0 \ \Omega is pivoted at one end and rotates with a constant angular velocity ω=10 rad/s\omega = 10 \text{ rad/s} in a horizontal plane. A uniform vertical magnetic field B=2.0 TB = 2.0 \text{ T} exists throughout the region. The free end of the rod slides on a fixed, perfectly conducting circular ring. The pivot and the ring are connected by a stationary wire containing an ideal battery of EMF V0=5.0 VV_0 = 5.0 \text{ V}. The battery is oriented such that it opposes the motional EMF generated by the rod. Let PBP_B be the net rate of work done by the magnetic force on all charge carriers in the rod, PemfP_{emf} be the rate at which energy is converted from mechanical to electrical form by the motional EMF, and PextP_{ext} be the mechanical power supplied by the external agent to maintain the constant angular velocity. Which of the following sets of values is correct?

Options

  • A.

    PB=0 W, Pemf=10 W, Pext=10 WP_B = 0 \text{ W}, \ P_{emf} = 10 \text{ W}, \ P_{ext} = 10 \text{ W}

  • B.

    PB=10 W, Pemf=10 W, Pext=10 WP_B = 10 \text{ W}, \ P_{emf} = 10 \text{ W}, \ P_{ext} = 10 \text{ W}

  • C.

    PB=0 W, Pemf=5.0 W, Pext=5.0 WP_B = 0 \text{ W}, \ P_{emf} = 5.0 \text{ W}, \ P_{ext} = 5.0 \text{ W}

  • D.

    PB=10 W, Pemf=5.0 W, Pext=10 WP_B = 10 \text{ W}, \ P_{emf} = 5.0 \text{ W}, \ P_{ext} = 10 \text{ W}

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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

Motional EMF in Rotating Rods|Conservative vs Non-conservative|Integration

Why This Question Is Tricky

Common Mistakes

This question targets a profound misconception regarding the work-energy theorem in electrodynamics.

Students frequently assume that the magnetic force does the work to drive the current, erroneously calculating PB=EI=10 WP_B = \mathcal{E}I = 10 \text{ W}.

In reality, the magnetic force strictly does zero net work as it is always perpendicular to the charge's velocity; it merely acts as a mediator, redirecting the mechanical work done by the external agent into electrical energy.

Furthermore, students may incorrectly calculate PemfP_{emf} or PextP_{ext} using the net EMF (5.0 V5.0 \text{ V}) rather than the motional EMF (10 V10 \text{ V}), failing to recognize that the external agent must supply enough power to overcome both the Joule heating (I2R=5.0 WI^2R = 5.0 \text{ W}) and the power absorbed by the opposing battery (V0I=5.0 WV_0 I = 5.0 \text{ W}).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptMotional EMF in Rotating Rods
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree