Let's go through each part of the problem.

Given Information

• Number of turns $N = 78$
• Dimensions of the coil $a = 2.50 \, \text{cm} = 0.025 \, \text{m}$ and $b = 4.00 \, \text{cm} = 0.04 \, \text{m}$
• Magnetic field $B = 0.800 \, \text{T}$
• Current $I = 10.8 \, \text{mA} = 0.0108 \, \text{A}$
• Rotational speed $f = 3600 \, \text{rev/min} = 60 \, \text{rev/sec}$

Let's proceed with each part:

Part (a): Maximum Torque Acting on the Rotor

The torque $\tau$ on a current-carrying coil in a magnetic field is given by: $\tau = N \cdot I \cdot A \cdot B \cdot \sin \theta$ where:

• $N$ is the number of turns,
• $I$ is the current,
• $A$ is the area of the coil $(A = a \cdot b)$,
• $B$ is the magnetic field strength,
• $\theta$ is the angle between the normal of the coil and the magnetic field.

For maximum torque, $\sin \theta = 1$, which happens when $\theta = 90^\circ$. Thus: $\tau_{\text{max}} = N \cdot I \cdot A \cdot B$ Calculating the area $A$: $A = a \cdot b = 0.025 \, \text{m} \times 0.04 \, \text{m} = 0.001 \, \text{m}^2$ Then, substituting values: $\tau_{\text{max}} = 78 \cdot 0.0108 \, \text{A} \cdot 0.001 \, \text{m}^2 \cdot 0.8 \, \text{T} = 0.00067392 \, \text{N} \cdot \text{m}$ So, the maximum torque $\tau_{\text{max}} \approx 0.000674 \, \text{N} \cdot \text{m}$.

Part (b): Peak Power Output of the Motor

The peak power output occurs when the rotor is moving through the position where