An aluminum ring of radius 5.00 cm and resistance 2.00 10-4 Ω is placed around the top of a long air-core solenoid with 997 turns per meter and a smaller radius of 3.00 cm, as in the figure below. If the current in the solenoid is increasing at a constant rate of 267 A/s, what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

Question

An aluminum ring of radius 5.00 cm and resistance 2.00  10-4 Ω is placed around the top of a long air-core solenoid with 997 turns per meter and a smaller radius of 3.00 cm, as in the figure below. If the current in the solenoid is increasing at a constant rate of 267 A/s, what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

Elena · Answer

The voltage induced around the Al ring is
ΔV=NΔΦ/Δt=Δ(0.5•B•A) /Δt =
=0.5•A •ΔB /Δt.

B =μ₀nI,
ΔB /Δt= Δ(μ₀•nvI)/ Δt,
ΔB /Δt= μ₀•n•ΔI/ Δt,
ΔV= 0.5•A •ΔB /Δt=
=0.5•A• μ₀•n•I ΔI/ Δt.

I= ΔV/R =
= (0.5•A• μ₀•n/R)• (ΔI/ Δt)=
=0.5•π•(0.03)² •4π•10⁻⁷•997•267/2•10⁻⁴=
=2.36 A.