alleryAt equilibrium, the deflection ϕ of a moving coil galvanometer with N turns, area A, current I, in a radial magnetic field B and spring constant k satisfies
Aϕ = NIAB/k
Bϕ = k/(NIAB)
Cϕ = NI/(kAB)
Dϕ = N²IAB/k
Answer & Solution
Correct answer: A. ϕ = NIAB/k
1. The magnetic torque on the coil is τ = NIAB (with the radial field design).
2. The restoring spring torque is k·ϕ where k is the torsional constant.
3. At equilibrium, the two torques balance: kϕ = NIAB.
4. Solve for ϕ: ϕ = NIAB/k. Deflection is linear in current.
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.11_
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