Final answer:
The torque on a current loop reaches 90.0% of maximum when the sin(θ) is 0.9, which yields an angle of approximately 64.16°, not matching the provided options. For 50.0% of maximum torque, the appropriate angle is 30°, whereas for 10.0% of maximum torque, the angle is approximately 5.74° or 174.26°.
Step-by-step explanation:
The torque τ on a current loop in a magnetic field is given by the equation τ = NIAB sin(θ), where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the normal to the loop and the magnetic field direction. The torque is at its maximum when θ is 90° because the sin(θ) is equal to 1. Therefore, since the question is seeking the angle for which the torque is 90.0% of maximum, we want to find the angle θ where sin(θ) equals 0.9, which corresponds to θ approximately 64.16°.
- In achieving 90.0% of maximum torque, the angle θ does not correspond to any of the options provided.
- For 50.0% of maximum torque, θ is 30° since sin(30°) = 0.5.
- For 10.0% of maximum torque, θ should be approximately 5.74° or 174.26° since sin(5.74°) ≈ 0.1.