A solenoid 1.5 m long and 3.0 cm in diameter carries a current of 12 A. The magn

Question

A solenoid 1.5 m long and 3.0 cm in diameter carries a current of 12 A. The magnetic field inside the solenoid is 3.3 mT. (a) How many turns of wire comprise the solenoid? (b) Find the length of wire used to make the solenoid. (c) What is the self-inductance of the solenoid? (d) If the current in the solenoid drops to zero in a time of 0.15 s, what is the induced emf in the solenoid? (e) What is the change in flux through the solenoid?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Explanation / Answer

a. B = uoNi/l

N = 0.0033* 1.5 /12.56*10^-7 * 12

N = 328 TURNS

B. 1 TURN HAS A LENGTH OF 2 * 3.14* 0.015 = 0.0942

TOTAL LENGTH l = 0.0942*328 = 30.89 m = 31 m

c.

L = u0 N^2 pi r^2/l find from this

d. emf e = L di/dt

e. chagne of flux = B A

kindly use those formulas and get the answers

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.