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Question #43357

Difficulty: 8/10 | 🟢 Free

The Question

Consider a rectangular primary loop ACDBACDB in the xyxy-plane with vertices A(0,0)A(0,0), B(L,0)B(L,0), D(L,w)D(L,w), and C(0,w)C(0,w). The side ABAB is a uniform potentiometer wire of length LL and resistance RR. The other sides are made of ideal resistanceless wires. An ideal primary cell of EMF V0V_0 is connected in the side CDCD such that it drives a steady current in the counter-clockwise direction. A uniform magnetic field B⃗=αtk^\vec{B} = \alpha t \hat{k} exists everywhere, where k^\hat{k} is the unit vector in the +z+z direction (out of the page) and α>0\alpha > 0 is a constant. A secondary circuit consists of an ideal cell of EMF EsE_s and a galvanometer. It connects point AA to a movable jockey JJ at position (x,0)(x,0) on the wire ABAB. The connecting wires of the secondary circuit form a smaller rectangular loop AJJ′C′A J J' C' with vertices A(0,0)A(0,0), J(x,0)J(x,0), J′(x,w/2)J'(x, w/2), and C′(0,w/2)C'(0, w/2). The cell EsE_s and the galvanometer are located in the branch C′AC'A, with the positive terminal of EsE_s connected to AA. Given V0>αLwV_0 > \alpha L w, find the balancing length xx for which the galvanometer shows zero deflection.

Options

  • A.

    2LEs2V0−αLw\frac{2 L E_s}{2 V_0 - \alpha L w}

  • B.

    LEsV0−αLw\frac{L E_s}{V_0 - \alpha L w}

  • C.

    2LEs2V0+αLw\frac{2 L E_s}{2 V_0 + \alpha L w}

  • D.

    LEsV0\frac{L E_s}{V_0}

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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

Potentiometer Sensitivity|AC Generator Back-EMF|Superposition of Magnetic Fields

Why This Question Is Tricky

Common Mistakes

This question tests the application of Faraday's Law in distributed circuits, challenging the misconception that Kirchhoff's Voltage Law (KVL) holds in time-varying magnetic fields.

Students often memorize the static potentiometer formula x∝Esx \propto E_s based on lumped potential drops.

The trap lies in failing to recognize that the induced electric field is non-conservative; the null point is a dynamic equilibrium governed by ∮E⃗⋅dl⃗=−dΦ/dt\oint \vec{E} \cdot d\vec{l} = -d\Phi/dt.

Common mistakes include accounting for the induced EMF in the primary loop but ignoring the flux enclosed by the secondary loop (Option B), or ignoring the primary loop's flux while accounting for the secondary's (Option C).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptPotentiometer Sensitivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree