A closed coil having 100 turns is rotated in a uniform magnetic field `B = 4.0 X 10^(-4)` T about a diameter which is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is `25 cm^2` and its resistance is `4.0 Omega.` Find (a) the average emf developed in the half a turn form a position where the coil is perpendicular to the magnetic field, (b) the average emf in a full turn and (c) the net charge displaced in part (a).

Explanation

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as requested.

### Given Data:
- Number of turns, \( N = 100 \)
- Magnetic field, \( B = 4.0 \times 10^{-4} \, \text{T} \)
- Area of the coil, \( A = 25 \, \text{cm}^2 = 25 \times 10^{-4} \, \text{m}^2 = 2.5 \times 10^{-3} \, \text{m}^2 \)
- Resistance, \( R = 4.0 \, \Omega \)
- Angular velocity, \( \omega = 300 \, \text{rev/min} \)
...