The rotor in a certain electric motor is a flat, rectangular coil with 70 turns of wire and dimensions 2.50 cm by 4.00 cm . The rotor rotates in a uniform magnetic field of 0.800 T . When the plane of the rotor is perpendicular to the direction of the magnetic field, it carries a current of 9.3 mA . In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at \( 3600 \mathrm{rev} / \mathrm{min} \). (a) Find the maximum torque acting on the rotor. \( \tau_{\text {max }}=5.208 \mathrm{e}-4 \) (b) Find the peak power output of the motor. \( \mathcal{P}_{\text {max }}=1.964 \mathrm{e}-1 \) (c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution. W \( =0.00327 \) Yourr response differs from the correct answer by more than \( 10 \% \). Double check your calculations. J (d) What is the average power of the motor? \( \mathcal{P} \) avg \( =1.964 \mathrm{e}-1 \) Your response differs from the correct answer by more than \( 10 \% \). Double check your calculations. W

To solve the problem, we will go through each part step by step. ### (a) Find the maximum torque acting on the rotor. The torque \(\tau\) on a current-carrying loop in a magnetic field is given by the formula: \[ \tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) \] where: - \(n\) = number of turns (70 turns) - \(I\) = current (9.3 mA = \(9.3 \times 10^{-3}\) A) - \(A\) = area of the coil - \(B\) = magnetic field (0.800 T) - \(\theta\) = angle between the magnetic moment and the magnetic field (90° for maximum torque) First, we need to calculate the area \(A\) of the rectangular coil: \[ A = \text{length} \times \text{width} = 0.025 \, \text{m} \times 0.040 \, \text{m} = 0.001 \, \text{m}^2 \] Now, substituting the values into the torque formula: \[ \tau_{\text{max}} = n \cdot I \cdot A \cdot B \] \[ \tau_{\text{max}} = 70 \cdot (9.3 \times 10^{-3}) \cdot (0.001) \cdot (0.800) \] \[ \tau_{\text{max}} = 70 \cdot 9.3 \times 10^{-3} \cdot 0.001 \cdot 0.800 \] \[ \tau_{\text{max}} = 70 \cdot 9.3 \times 0.0008 \times 10^{-3} \] \[ \tau_{\text{max}} = 70 \cdot 0.00000744 \] \[ \tau_{\text{max}} = 5.208 \times 10^{-4} \, \text{N m} \] ### (b) Find the peak power output of the motor. The peak power \(P\) can be calculated using the formula: \[ P = \tau \cdot \omega \] where \(\omega\) is the angular velocity in radians per second. To convert the rotational speed from revolutions per minute (rev/min) to radians per second: \[ \omega = 3600 \, \text{rev/min} \cdot \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \cdot \frac{1 \, \text{min}}{60 \, \text{s}} = 3600 \cdot \frac{2\pi}{60} = 376.99 \, \text{rad/s} \] Now, substituting the values into the power formula: \[ P_{\text{max}} = \tau_{\text{max}} \cdot \omega \] \[ P_{\text{max}} = (5.208 \times 10^{-4}) \cdot (376.99) \] \[ P_{\text{max}} \approx 1.964 \times 10^{-1} \, \text{W} \] ### (c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution. The work done \(W\) in one full revolution can be calculated as: \[ W = \tau_{\text{max}} \cdot \theta \] where \(\theta\) is the angle in radians for one full revolution (which is \(2\pi\) radians). \[ W = \tau_{\text{max}} \cdot 2\pi \] \[ W = (5.208 \times 10^{-4}) \cdot (2\pi) \] \[ W \approx (5.208 \times 10^{-4}) \cdot 6.2832 \approx 0.00327 \, \text{J} \] ### (d) What is the average power of the motor? The average power can be calculated as the work done per unit time. Since the rotor completes one full revolution in the time it takes to make 3600 revolutions per minute: \[ \text{Time for one revolution} = \frac{1 \, \text{min}}{3600 \, \text{rev}} = \frac{60 \, \text{s}}{3600} = \frac{1}{60} \, \text{s} \] Now, the average power \(P_{\text{avg}}\) is: \[ P_{\text{avg}} = \frac{W}{\text{Time for one revolution}} = \frac{0.00327}{\frac{1}{60}} = 0.00327 \times 60 \approx 0.1964 \, \text{W} \] Thus, the average power is approximately: \[ P_{\text{avg}} \approx 1.964 \times 10^{-1} \, \text{W} \] ### Summary of Results: - (a) \(\tau_{\text{max}} = 5.208 \times 10^{-4} \, \text{N m}\) - (b) \(P_{\text{max}} = 1.964 \times 10^{-1} \, \text{W}\) - (c) \(W = 0.00327 \, \text{J}\) - (d) \(P_{\text{avg}} = 1.964 \times 10^{-1} \, \text{W}\) If your responses differ from the expected answers, please double-check the calculations and ensure that all units are consistent.