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Question #43931

Difficulty: 8/10 | 🟢 Free

The Question

A circular conducting loop of radius aa (in m\text{m}), self-inductance LL (in H\text{H}), and negligible ohmic resistance is placed in a uniform magnetic field BB (in T\text{T}) perpendicular to its plane. The loop is connected in series with an electrolytic cell of negligible EMF. The initial internal resistance of the cell is R0R_0 (in Ω\Omega). At t=0t=0, the magnetic field is suddenly switched off. As the induced transient current flows, the electrolyte heats up, causing its internal resistance to decrease exponentially with the total charge qq (in C\text{C}) that has passed through it since t=0t=0, given by R(q)=R0e−λqR(q) = R_0 e^{-\lambda q}, where λ\lambda is a positive constant in C−1\text{C}^{-1}. Assuming λBπa2<R0\lambda B \pi a^2 < R_0, what is the total charge QQ (in C\text{C}) that flows through the circuit until the current completely dies out?

Options

  • A.

    −1λln⁡(1−λBπa2R0)-\frac{1}{\lambda} \ln\left( 1 - \frac{\lambda B \pi a^2}{R_0} \right)

  • B.

    Bπa2R0\frac{B \pi a^2}{R_0}

  • C.

    1λln⁡(1+λBπa2R0)\frac{1}{\lambda} \ln\left( 1 + \frac{\lambda B \pi a^2}{R_0} \right)

  • D.

    Bπa2R0e−λBπa2/R0\frac{B \pi a^2}{R_0} e^{-\lambda B \pi a^2 / R_0}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Cell EMF vs Terminal Voltage|Transient Current in RL Circuit|Integration

Why This Question Is Tricky

Common Mistakes

Students commonly misapply the steady-state charge formula Q=ΔΦRQ = \frac{\Delta \Phi}{R} for electromagnetic induction, failing to recognize that it is only valid when the circuit resistance is constant.

In this problem, the internal resistance of the cell is dynamically coupled to the transient current, decreasing as charge flows.

The primary trap (Option B) catches students who blindly use the initial resistance R0R_0 and ignore the non-linear nature of the RL circuit.

The correct physical framework requires applying Faraday's Law in its fundamental integral form, ∫Edt=∫IR(q)dt+∫LdI\int \mathcal{E} dt = \int I R(q) dt + \int L dI, and recognizing that the resistive dissipation term transforms into an integral over charge, ∫R(q)dq\int R(q) dq, allowing for an exact analytical solution despite the non-linear coupling.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptCell EMF vs Terminal Voltage
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree