A wire coil of 20 turns has a cross-sectional area of 16 cm2. The coil is placed

Question

A wire coil of 20 turns has a cross-sectional area of 16 cm2. The coil is placed in a uniform field of a large magnet with B = 0.12 T. The coil is suddenly rotated 90 o from an orientation parallel to the field to one perpendicular to the field. The time to flip the coil is 0.65 s.

a) What is the magnitude of the average emf produced?

b) What will be the average current produced if the circuit of the coil is closed and the resistance is 80 ohm?

Please show all work! I will rate very well! Thanks!

Explanation / Answer

e =NAdB/dt =20*16*10-4*0.12/0.65 = 5.9 mV

I=e/R = 7.38 x 10-5 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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