Chapter 21, problem 43partially correct answer. your answer is partially correct. try again. the 1400-turn coil in a dc motor has an area per turn of 1.1 × 10-2 m2. the design for the motor specifies that the magnitude of the maximum torque is 8.4 n·m when the coil is placed in a 0.18-t magnetic field. what is the current in the coil?

I = 3.028 A

Explanation:

given,

Number of turns in the coil = N = 1400

Area of DC motor per turn = 1.1 × 10⁻² m²

Maximum torque in motor = 8.4 N.m

Magnetic field in the coil = 0.18 T

Current in the coil = ?

using formula for toque in the magnetic field

   τ = N A B I

   8.4 = 1400 x 1.1 × 10⁻²  x 0.18 x I

         I = 3.028 A

Current in the coil will be equal to I = 3.028 A

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s