The figure shows a cross section of three parallel wires each carrying a current

Question

The figure shows a cross section of three parallel wires each carrying a current of 20 A. The currents in wires A and B are out of the paper, while that in wire C is into the paper. If the distance R = 5.0 mm, what is the a magnitude of the force on a 0.5-m length of wire A? A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 1.0 s. If the coil has a total resistance of 0.20ohm. what is the magnitude of the induced current?

Explanation / Answer

1. 23 mN

2, induced emf e = NADB/dt

E = 20 * 50e-4* (6-2)/1

e = 0.4 V

induced current i = emfc/R = 0.4/0.2 = 2 A

option B iti s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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