Answer:
(a) τ = 0.782Nm
(b) x = 0.747Nm/rad
Step-by-step explanation:
Given that
N = 50, r = 1.75cm = 1.75×10-²m, B = 0.25T, I = 65A.
The torque on a coil of N turns of wire is given by
τ = NIAB
From the problem statement, this torque can also be represented by an angular form of hooke's law.
τ = xθ
Where N = number of turns, I = current in amps (A), A = area in m², B = magnetic field strength in T, x = torque constant in Nm/rad and θ = angular displacement.
Area A = πr² = π×(1.75×10-²)² = 9.62×10-⁴m²
(a) τ = NIAB = 50×65×9.62×10-⁴×0.25
τ = 0.782Nm
(b) x = τ/θ
θ = 60° = 60×2π/360 = π/3 rad
x = 0.782/(π/3)
x = 0.747Nm/rad.