COLLEGE PHYSICS V1+WEBASSIGN MULTI-TERM
COLLEGE PHYSICS V1+WEBASSIGN MULTI-TERM
11th Edition
ISBN: 9780357683538
Author: SERWAY
Publisher: CENGAGE L
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Chapter 20, Problem 57AP

An 820-turn wire coil of resistance 24.0 Ω is placed on lop of a 12 500-turn, 7.00-cm-long solenoid, as in Figure P20.57. Both coil and solenoid have cross-sectional area of 1.00 × 10−4 m2. (a) How long does it take the solenoid current to reach 0.632 times its maximum value? (b) Determine the average back emf caused by the self-inductance of the solenoid during this interval. The magnetic field produced by the solenoid at the location of the coil is one-half as strong as the field at the center of the solenoid. (c) Determine the average rate of change in magnetic flux through each turn of the coil during the stated interval. (d) Find the magnitude of the average induced current in the coil.

Chapter 20, Problem 57AP, An 820-turn wire coil of resistance 24.0  is placed on lop of a 12 500-turn, 7.00-cm-long solenoid,

Figure P20.57

(a)

Expert Solution
Check Mark
To determine
The time taken by the solenoid current to reach 0.632 times its maximum value.

Answer to Problem 57AP

Solution: The time taken by the solenoid current to reach 0.632 times its maximum value is 20.0ms .

Explanation of Solution

Given Info: The number of turns is 12500, cross sectional area of the solenoid is 1.00×104m2 , and length of the solenoid is 7.00cm .

Formula to calculate the self-inductance of the solenoid coil is,

L=μ0N2Al

  • L is the self-inductance,
  • μ0 is the permeability of free space,
  • N is the number of turns,
  • A is the cross sectional area,
  • l is the length of the solenoid,

Substitute 12500-turn for N, 4π×107Tm/A for μ0 , 1.00×104m2 for A and 7.00cm for l.

L=(4π×107Tm/A)(12500)2(1.00×104m2)(7.00cm)(102m1cm)=0.280H

Thus, the self-inductance of the solenoid coil is 0.280 H.

Conclusion:

Since the current in the solenoid reaches a maximum current of Is=0.632Imax in a time of t=τ , so that,

The formula to calculate the time period is,

t=LR

  • t is the time taken by the solenoid current to reach 0.632 times its maximum value,
  • R is the resistance of the circuit,

Substitute 0.280 H for L and 14.0Ω for R.

t=(0.280H)(14.0Ω)=0.02s=2.00×102s20.0ms

Therefore, the time taken by the solenoid current to reach 0.632 times its maximum value is 20.0ms .

(b)

Expert Solution
Check Mark
To determine
The average back emf caused by the self-inductance of the solenoid.

Answer to Problem 57AP

Solution: The average back emf caused by the self-inductance of the solenoid is

37.9 V.

Explanation of Solution

Given Info: The self-inductance of the solenoid coil is 0.280 H, potential difference across the battery is 60.0V , and resistance of the circuit is 14.0Ω .

Formula to calculate the change in the solenoid current at the 0.632 times is,

ΔIs=0.632Imax0

  • ΔIs is the change in the solenoid current,
  • Imax is the maximum value of the current,

Use ΔV/R for Imax in the above relation.

ΔIs=0.632(ΔVR)

  • ΔV is the potential difference across the battery,

Substitute 60.0V for ΔV and 14.0Ω for R to find ΔIs .

ΔIs=0.632(60.0V14.0Ω)=2.71A

Thus, the change in the solenoid current is 2.71 A.

Conclusion:

Formula to calculate the average back emf is,

εback=L(ΔIsΔt)

  • εback is the back emf,
  • L is the self-inductance,
  • Δt is the time taken by the solenoid current,

Substitute 0.280H for L, 2.71A for ΔIs , and 20.0ms for Δt .

εback=(0.280H)[(2.71A)(20.0ms)(103s1ms)]=0.0379×103V=37.9V

Therefore, the average back emf caused by the self-inductance of the solenoid is

37.9 V.

(c)

Expert Solution
Check Mark
To determine
The average rate of change in magnetic flux through each turn of the coil.

Answer to Problem 57AP

Solution: The rate of change in magnetic flux through each turn of the coil is 1.52×103V .

Explanation of Solution

Given Info: The number of turns is 12500, cross sectional area of the solenoid is 1.00×104m2 , the time taken by the solenoid current to reach 0.632 times its maximum value is 2.00×102s and length of the solenoid is 7.00cm .

Since the change in the magnetic field at the location of the coil is one half of the change in the field at the center of the solenoid, so that ΔBcoil=12[μ0ns(ΔIs)] .

Formula to calculate the average rate of change of flux through each turn of the coil is, ΔΦBΔt=(ΔBcoil)AcoilΔt

  • ΔΦB/Δt is the rate of change of flux,
  • ΔBcoil/Δt is the rate of change of magnetic field of the coil,
  • Acoil is the cross section area of the coil,

Use 1/2[μ0ns(ΔIs)] for ΔBcoil and Ns/ls for ns in the above relation,

ΔΦBΔt=12[μ0Ns(ΔIs)]AcoillsΔt=μ0Ns(ΔIs)Acoil2ls(Δt)

  • μ0 is the permeability of free space,
  • Ns is the number of turns in the solenoid,
  • ΔIs is the is the change in the solenoid current,
  • ls is the length of the solenoid,

Substitute 4π×107Tm/A for μ0 , 12500 for Ns , 2.71A for ΔIs , 1.00×104m2 for Acoil , 7.00cm for ls , and 2.00×102s for Δt .

ΔΦBΔt=(4π×107Tm/A)(12500)(2.71A)(1.00×104m2)2(7.00cm)(102m1cm)(2.00×102s)=1.52×103V

Conclusion:

Therefore, the rate of change in magnetic flux through each turn of the coil is 1.52×103V .

The number of turns is 12500, cross sectional area of the solenoid is 1.00×104m2 , the time taken by the solenoid current to reach 0.632 times its maximum value is 2.00×102s and length of the solenoid is 7.00cm .

Since the change in the magnetic field at the location of the coil is one half of the change in the field at the center of the solenoid, so that ΔBcoil=12[μ0ns(ΔIs)] .

Formula to calculate the average rate of change of flux through each turn of the coil is, ΔΦBΔt=(ΔBcoil)AcoilΔt

  • ΔΦB/Δt is the rate of change of flux,
  • ΔBcoil/Δt is the rate of change of magnetic field of the coil,
  • Acoil is the cross section area of the coil,

Use 1/2[μ0ns(ΔIs)] for ΔBcoil and Ns/ls for ns in the above relation,

ΔΦBΔt=12[μ0Ns(ΔIs)]AcoillsΔt=μ0Ns(ΔIs)Acoil2ls(Δt)

  • μ0 is the permeability of free space,
  • Ns is the number of turns in the solenoid,
  • ΔIs is the is the change in the solenoid current,
  • ls is the length of the solenoid,

Substitute 4π×107Tm/A for μ0 , 12500 for Ns , 2.71A for ΔIs , 1.00×104m2 for Acoil , 7.00cm for ls , and 2.00×102s for Δt .

ΔΦBΔt=(4π×107Tm/A)(12500)(2.71A)(1.00×104m2)2(7.00cm)(102m1cm)(2.00×102s)=1.52×103V

Conclusion:

Therefore, the rate of change in magnetic flux through each turn of the coil is 1.52×103V .

(d)

Expert Solution
Check Mark
To determine
The magnitude of the average induced current in the coil.

Answer to Problem 57AP

Solution: The magnitude of the average induced current in the coil is 51.9mA .

Explanation of Solution

Given Info: The rate of change in magnetic flux through each turn of the coil is 1.52×103V number of turns in the coil is 820, and resistance of the coil is 24.0Ω .

Formula to calculate the average induced current in the coil is,

Icoil=εcoilRcoil

  • Icoil is the induced current in the coil,
  • εcoil is the emf across the coil,
  • Rcoil is the resistance of the coil,

Use Ncoil(ΔΦB/Δt)coil for εcoil in the above relation.

Icoil=Ncoil(ΔΦB/Δt)coilRcoil

  • Ncoil is the number of turns in coil,

Substitute 820 turns for Ncoil , 1.52×103V for (ΔΦB/Δt)coil , and 24.0Ω for Rcoil .

Icoil=(820)(1.52×103V)(24.0Ω)=51.9×103A=51.9mA

Conclusion:

Therefore, the magnitude of the average induced current in the coil is 51.9mA .

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Chapter 20 Solutions

COLLEGE PHYSICS V1+WEBASSIGN MULTI-TERM

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