Consider a 50-turn circular loop with a radius of 1.75 cm in a 0.25-T magnetic field. This coil is going to be used in a galvanometer that reads 65 for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is Here x is the torque constant and? is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel? 50%

Part (a) Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it

Part (b) What is the torque constant of the spring, in newton meters per radian, that must be used in this device?

Question

Physics

Alayna Lloyd

3 June, 21:47

Consider a 50-turn circular loop with a radius of 1.75 cm in a 0.25-T magnetic field. This coil is going to be used in a galvanometer that reads 65 for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is Here x is the torque constant and? is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel? 50%

Part (a) Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it

Part (b) What is the torque constant of the spring, in newton meters per radian, that must be used in this device?

+1

Answers (1)

Know the Answer?

Adriana Kirk · Answer 1

(a) τ = 0.782Nm

(b) x = 0.747Nm/rad

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.