An aluminum ring with a radius of 4.00 cm and a resistance 1.50 x 10-4 is placed

Question

An aluminum ring with a radius of 4.00 cm and a resistance 1.50 x 10-4 is placed on top of a long air core solenoid with 500 turns per meter and a radius of 1.50 cm, as shown in the figure below. Assume that the axial component of the field produced by the solenoid over the area of the end of the solenoid is one-half as strong as at the center of the solenoid, (a) If the current in the solenoid is increasing at a rate of 250 A/s, what is the induced current in the ring ? (b) At the center of the ring, what is the magnetic field produced by the induced current in the ring ? (c) What is the direction of this field ?

Explanation / Answer

Idea:

The magnitude of induced emf is || = d(BA)/dt

                                                       = 0.500nA(dI/dt)

                                                       = 0.50(4*10-7 Tm/A)(500)(*r22)*(dI/dt)

                                                        = 0.50(4*10-7 Tm/A)(500)(*(0.015 m)2)*(250 A/s)

                                                        = 554.6025*10-7 V

Solve:

a) the induced current in the ring is

   I=||/R

    = 554.6025*10-7 V/1.50*10-4

    = 0.369735 A

b)

The magnetic field produced due to the induced current in the ring is

Bring = 0I/2r1

   = (4*10-7 Tm/A)( 0.369735 A)/2*(0.04m)

   = 5.8048395*10-6 T

c)

Coils field points downward and is increasing, so Bring points upward

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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