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Question #42740

Difficulty: 8/10 | 🟢 Free

The Question

A rigid rectangular wire loop is placed in a uniform magnetic field B⃗\vec{B} perpendicular to its plane, pointing into the page. The loop is divided into two identical adjacent square loops, Loop 1 (left) and Loop 2 (right), by a central straight wire. The side length of each square is aa (in m\text{m}). The magnetic field magnitude increases with time as B(t)=βtB(t) = \beta t, where β>0\beta > 0 is a constant in T/s\text{T/s}. The left-most vertical branch has a resistance RR (in Ω\Omega), and the right-most vertical branch has a resistance 2R2R (in Ω\Omega). The top and bottom horizontal branches have negligible resistance. The central vertical branch contains an ideal inductor of inductance LL (in H\text{H}) and has negligible resistance. Assume the self-inductance of the square loops is negligible compared to LL. At t=0t=0, the magnetic field starts increasing. Find the magnitude of the current flowing through the central branch as a function of time tt (in s\text{s}).

Options

  • A.

    a2β2R(1−e−2R3Lt) A\frac{a^2 \beta}{2R} \left( 1 - e^{-\frac{2R}{3L} t} \right) \text{ A}

  • B.

    3a2β2R(1−e−2R3Lt) A\frac{3a^2 \beta}{2R} \left( 1 - e^{-\frac{2R}{3L} t} \right) \text{ A}

  • C.

    a2βR(1−e−RLt) A\frac{a^2 \beta}{R} \left( 1 - e^{-\frac{R}{L} t} \right) \text{ A}

  • D.

    a2β3R(1−e−3RLt) A\frac{a^2 \beta}{3R} \left( 1 - e^{-\frac{3R}{L} t} \right) \text{ A}

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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

Kirchhoff’s Junction Rule|Transient Current in RL Circuit|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Kirchhoff's Loop Rule in the presence of non-conservative electric fields.

Students frequently misapply the concept of scalar potential by treating the induced EMFs as lumped batteries that both push current in the same direction through the shared central branch (leading to the incorrect option B).

The trap exploits the failure to recognize that the CCW induced electric field in the right loop actually opposes the CCW field of the left loop within the central branch.

By forcing the transition from algebraic Kirchhoff laws to a system of differential equations governed by Faraday's Law, it rigorously evaluates whether the student correctly integrates the non-conservative field path-dependently.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptKirchhoff’s Junction Rule
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree