Question

The 1400-turn coil in a dc motor has an area per turn of 1.2 × 10^-2 m². The design for the motor specifies that the magnitude of the maximum torque is 3.5 N·m when the coil is placed in a 0.23-T magnetic field. What is the current in the coil?

Answer 1

$$\tau_{max}=NIABsin\theta,$$ $$I=\dfrac{\tau_{max}}{NABsin\theta},$$ $$I=\dfrac{3.5\ N\cdot m}{1400\cdot1.2\cdot10^{-2}\ m^2\cdot0.23\ T\cdot sin90^{\circ}}=0.9\ A.$$