EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
10th Edition
ISBN: 8220106906149
Author: Jewett
Publisher: Cengage Learning US
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Chapter 30, Problem 48CP

An induction furnace uses electromagnetic induction to produce eddy currents in a conductor, thereby raising the conductor’s temperature. Commercial units operate at frequencies ranging from 60 Hz to about 1 MHz and deliver powers from a few watts to several megawatts. Induction heating can be used for warming a metal pan on a kitchen stove. It can be used to avoid oxidation and contamination of the metal when welding in a vacuum enclosure. To explore induction heating, consider a flat conducting disk of radius R, thickness b, and resistivity ρ. A sinusoidal magnetic field Bmax cos ωt is applied perpendicular to the disk. Assume the eddy currents occur in circles concentric with the disk. (a) Calculate the average power delivered to the disk. (b) What If? By what factor does the power change when the amplitude of the field doubles? (c) When the frequency doubles? (d) When the radius of the disk doubles?

(a)

Expert Solution
Check Mark
To determine
The average power delivered to the disk.

Answer to Problem 48CP

The average power delivered to the disk is πR4ω2B2maxb16ρ .

Explanation of Solution

Given info: Radius of disk is R , thickness of disk is b , resistivity of the disk is ρ and sinusoidal magnetic field is Bmaxcosωt .

Since the eddy currents occur as concentric circles with the disk. Consider the disk to be a collection of rings that each has an induced emf.

The emf induced in the disk can be given as,

ε=d(BA)dt

Here,

ε is the emf induced in the disk.

B is the magnetic field.

A is the area of the disk.

Substitute Bmaxcosωt for B and πr2 for A in the above equation,

ε=d(Bmaxcosωt)(πr2)dt=πr2ωBmaxsinωt

Here,

Bmax is the maximum magnetic field.

ω is the angular velocity.

r is the radius of the disk.

t is the time period of the disk.

The elemental resistance around the ring can be given as,

dR=ρlringAring

Here,

R is resistance in the ring

ρ is the resistivity of the disk

lring is the length of the elemental ring

Aring is the area of the elemental ring

Substitute 2πr for lring and bdr for Aring in the above equation,

dR=ρ(2πr)bdr

The power delivered to the elemental ring can be given as,

dP=ε2dR

P is the power delivered to the ring,

Substitute ρ(2πr)bdr for dR and πr2ωBmaxsinωt for ε in the above equation,

dP=(πr2ωBmaxsinωt)2(ρ(2πr)bdr)=πr3ω2B2maxsin2(ωt)bdr2ρ

The total power delivered to the disk can be given as,

P=dP

Substitute πr3ω2B2maxsin2(ωt)bdr2ρ for dP in the above equation,

P=0Rπr3ω2B2maxsin2(ωt)bdr2ρ=πω2B2maxbsin2(ωt)2ρ0Rr3dr=πR4ω2B2maxsin2(ωt)bdr8ρ

Substitute 12 for sin2(ωt) to get average power in the above equation,

Pavg=πR4ω2B2maxb16ρ (1)

Here,

Pavg is the average power delivered to the disk.

Thus, the average power delivered to the disk can be given as πR4ω2B2maxb16ρ .

Conclusion:

Therefore, the average power delivered to the disk can be given as πR4ω2B2maxb16ρ .

(b)

Expert Solution
Check Mark
To determine
The factor by which power will change when the field doubles.

Answer to Problem 48CP

Answer The factor by which power will change when the field doubles is four times.

Explanation of Solution

Given info: Radius of disk is R , thickness of disk is b , resistivity of the disk is ρ and sinusoidal magnetic field is Bmaxcosωt .

Explanation:

The relation between the field and the power can be given from equation (1) as,

PavgB2max

Substitute 2Bmax for Bmax in the above equation,

Pavg(2B)2max4B2max4Pavg

Thus, the power will change by four times when the field doubles.

Conclusion:

Therefore, the factor by which power will change when the field doubles is four times.

(c)

Expert Solution
Check Mark
To determine
The factor by which power will change when the frequency doubles.

Answer to Problem 48CP

Answer The factor by which power will change when the frequency doubles is four times.

Explanation of Solution

Given info: Radius of disk is R , thickness of disk is b , resistivity of the disk is ρ and sinusoidal magnetic field is Bmaxcosωt .

Explanation:

The relation between the field and the power can be given from equation (1) as,

Pavgω2

Substitute 2πf for ω in the above equation,

Pavg(2πf)2f2

Here,

f is the frequency of the disk.

Substitute 2f for f in the above equation,

Pavg(2f)24f24Pavg

Thus, the power will change by four times when the frequency doubles.

Conclusion:

Therefore, the factor by which power will change when the frequency doubles is four times.

(d)

Expert Solution
Check Mark
To determine
The factor by which power will change when the radius of the disk doubles.

Answer to Problem 48CP

Answer The factor by which power will change when the radius of the disk doubles is sixteen times.

Explanation of Solution

Given info: Radius of disk is R , thickness of disk is b , resistivity of the disk is ρ and sinusoidal magnetic field is Bmaxcosωt .

Explanation:

The relation between the field and the power can be given from equation (1) as,

PavgR4

Substitute 2R for R in the above equation,

Pavg(2R)416R416Pavg

Thus, the power will change by sixteen times when the radius of disk doubles.

Conclusion:

Therefore, the factor by which power will change when the radius of disk doubles is sixteen times.

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Chapter 30 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY