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Question #42979

Difficulty: 8/10 | 🟢 Free

The Question

A stationary loop of inductance L=2.0 HL = 2.0 \text{ H} and zero resistance is placed in a uniform magnetic field. The initial magnetic flux through the loop is Φ0=20 Wb\Phi_0 = 20 \text{ Wb}. A thermally insulated resistor is connected in series with the loop, but is initially short-circuited by a superconducting switch. At t=0t=0, the external magnetic field is abruptly reduced to zero. Immediately after, the switch is opened so that the induced current flows through the resistor. The resistor has an initial resistance R0=1.0 ΩR_0 = 1.0 \ \Omega and a constant heat capacity C=5.0 J/KC = 5.0 \text{ J/K}. Its resistance varies with its temperature TT as R(T)=R0[1+α(T−T0)]R(T) = R_0 [1 + \alpha (T - T_0)], where T0T_0 is the initial temperature and α=0.05 K−1\alpha = 0.05 \text{ K}^{-1}. Assuming no heat is lost to the surroundings, find the time tt it takes for the current to drop to half of its initial value.

Options

  • A.

    0.5ln⁡(7) s0.5 \ln(7) \text{ s}

  • B.

    2.0ln⁡(2) s2.0 \ln(2) \text{ s}

  • C.

    1.0ln⁡(7) s1.0 \ln(7) \text{ s}

  • D.

    0.5ln⁡(5) s0.5 \ln(5) \text{ s}

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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

Faraday’s Law Vector Form|Energy Stored in Inductor|Internal Resistance Variation|First Law Application

Why This Question Is Tricky

Common Mistakes

This question tests the coupling between Faraday's Law and the First Law of Thermodynamics.

Students commonly fall into the trap of treating the resistance RR as a constant, blindly applying the standard L/RL/R time constant formula I(t)=I0e−Rt/LI(t) = I_0 e^{-Rt/L} to get t=(L/R0)ln⁡(2)t = (L/R_0)\ln(2).

This fundamentally ignores the feedback loop inherent in the system: the induced current dissipates thermal energy via Joule heating, which dynamically increases the internal resistance, thereby accelerating the decay of the current.

Solving it requires setting up a non-linear differential equation where the mechanical/electrical work done by the inductor is directly linked to the thermodynamic state of the resistor.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectromagnetic Induction - Class 12
ConceptFaraday’s Law Vector Form
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree