A solenoid has a cross-sectional area of 6.0 3 1024 m2, 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-V 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.

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