A 100-turn square loop 10.0 cm on a side carries 15.0 A in a 2.00 T field. What is the maximum torque on it?

1. Use torque = NIAB sin(theta), which is largest when sin(theta) = 1. 2. So the maximum torque is NIAB. 3. Work out the area: a square 10.0 cm on a side is 0.100 m on a side, so A = 0.0100 m^2. 4. Substitute: torque = (100)(15.0 A)(0.0100 m^2)(2.00 T). 5. Multiply step by step: (100)(15.0) = 1500 A, then (1500)(0.0100) = 15.0 A m^2. 6. Finally (15.0 A m^2)(2.00 T) = 30.0 N m. 7. So the maximum torque is 30.0 N m, large enough to be useful in a motor. 8. An answer of 300 N m comes from using 0.100 m^2 for the area, which is the area of a loop 31.6 cm on a side. _Source: OpenStax College Physics (CC BY 4.0), Ch 22 "Magnetism", section 22.8 Torque on a Current Loop: Motors and Meters_