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Question #43416

Difficulty: 8/10 | 🟢 Free

The Question

A moving coil galvanometer has a coil resistance Rg=18 ΩR_g = 18\ \Omega, N=100N = 100 turns, area A=10−3 m2A = 10^{-3}\ \text{m}^2, and is placed in a radial magnetic field B=0.2 TB = 0.2\ \text{T}. It is converted into an ammeter using a shunt resistance Rs=2 ΩR_s = 2\ \Omega. The coil and pointer assembly has a total mass m=20 gm = 20\ \text{g} with its center of mass located at a distance d=1 cmd = 1\ \text{cm} from the vertical pivot axis. The pivot has a torsional spring constant k=0.01 N⋅m/radk = 0.01\ \text{N}\cdot\text{m/rad}. The entire ammeter is mounted on a horizontal centrifuge rotating at a constant angular velocity ω=10 rad/s\omega = 10\ \text{rad/s} about a vertical axis. The galvanometer's pivot axis is at a distance L=0.5 mL = 0.5\ \text{m} from the centrifuge axis. When no current flows, the torsional spring is relaxed and the pointer's center of mass is aligned exactly radially outward from the centrifuge axis. If a steady total current I=1.0 AI = 1.0\ \text{A} is passed through the ammeter, find the steady angular deflection θ\theta of the pointer. (Assume θ\theta is small so that sin⁡θ≈θ\sin\theta \approx \theta).

Options

  • A.

    0.10 rad0.10\ \text{rad}

  • B.

    0.20 rad0.20\ \text{rad}

  • C.

    0.05 rad0.05\ \text{rad}

  • D.

    0.15 rad0.15\ \text{rad}

Check Your Answer

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

Shunt Resistance Conversion|Frame of Reference Inertial vs Non-Inertial|Pseudo Force Directionality

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to bridge electrodynamics with rigid body dynamics in a non-inertial frame.

The primary trap lies in calculating the restoring torque.

Students commonly calculate the magnetic torque correctly but equate it only to the mechanical spring torque (kθk\theta), completely ignoring the pseudo-force's contribution.

By placing the center of mass radially outward, the centrifugal force creates a stable equilibrium potential well.

Using the principle of virtual work (dU=−τdθdU = -\tau d\theta), the problem elegantly bypasses complex integration of differential centrifugal forces, revealing that the effective torsional constant increases by mLdω2mLd\omega^2.

Failing to account for this frame-dependent stress leads directly to the distractor option of 0.20 rad.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptShunt Resistance Conversion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree