Complete Question:
A solenoid has a cross-sectional area of 6.0×10−4m2 consists of 400 turns per meter, and carries a current of 0.40 A. A 10-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 1.5−Ω resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.
Answer:
I = 1.65 * 10⁻⁵A
Explanation:
I = Absolute value[-N(d∅/dt)/R]
N = 10 turns
∅ = BA cos Θ
B = μηI
B = (4π * 10⁻⁷)(400)(0.4)
B = 0.0002 T
A = 6.03 * 10⁻⁴m²
∅ = 0.0002 * 6.03 * 10⁻⁴
∅ = 12.12 * 10⁻⁸
d∅/dt = (12.12 * 10⁻⁸)/0.05
d∅/dt = 24.24 * 10⁻⁷
I =Abslute value of [-10 * ( 24.24 * 10⁻⁷)/1.5]
I = 1.6 * 10⁻⁵A
