COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 20, Problem 20QAP
To determine

(a)

Induced emf at t < 0 s

Expert Solution
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Answer to Problem 20QAP

Induced emf at t < 0 s= 0 T

Explanation of Solution

Given info:

  Number of turns =30diameter of the coil =0.6 mmagnetic field = 1.0 T

Formula used:

  A=πr2A=arear=radiusΦ=NBAΦ=magnetic fluxN=number of turnsε=dΦdtε= emft= time

Calculation:

  radius=0.6 m2=0.3 m

  By substituting,ε=dΦdtε=ddt(NBπr2)ε=Nπr2dBdtε=-Nπr2(B2B1)dtMagnetic field is constant when t < 0 sdB = 0,ε=0

Conclusion:

Induced emf at t < 0 s= 0 T

To determine

(b)

Induced emf at t =5.0 s

Expert Solution
Check Mark

Answer to Problem 20QAP

Induced emf at t =5.0 s= 2.54 mV

Explanation of Solution

Given info:

  Number of turns =30diameter of the coil =0.06 mmagnetic field = 1.0 Tmagnetic field is increased to 1.3 T in 10 s

Formula used:

  A=πr2A=arear=radiusΦ=NBAN=number of turnsε=dΦdtε= emft= time

Calculation:

  radius=0.06 m2=0.03 mMagnetic field at 5 s = 1.3 T - 1.0 T10 s*5 s+1.0 T=1.15 T

  By substituting,ε=dΦdtε=ddt(NBπr2)ε=Nπr2dBdtε=-Nπr2(1.15 T1.00 T)dtε=-30*3.14*(0.03 m)2(1.15 T1.00 T)1 sε=2.54 mV

Conclusion:

Induced emf at t =5.0 s= 2.54 mV

To determine

(c)

Induced emf at t < 10 s

Expert Solution
Check Mark

Answer to Problem 20QAP

Induced emf at t < 10 s= 0 T

Explanation of Solution

Given info:

  Number of turns =30diameter of the coil =0.6 mmagnetic field = 1.0 T

Formula used:

  A=πr2A=arear=radiusΦ=NBAN=number of turnsε=dΦdtε= emft= time

Calculation:

  radius=0.6 m2=0.3 m

  By substituting,ε=dΦdtε=ddt(NBπr2)ε=Nπr2dBdtε=-Nπr2(B2B1)dtMagnetic field is constant after 10 sdB = 0,ε=0

Conclusion:

Induced emf at t < 10 s= 0 T

To determine

(d)

Plot the magnetic field and induced emf as functions of time

Expert Solution
Check Mark

Answer to Problem 20QAP

  COLLEGE PHYSICS, Chapter 20, Problem 20QAP , additional homework tip  1

  COLLEGE PHYSICS, Chapter 20, Problem 20QAP , additional homework tip  2

Explanation of Solution

Given info:

    Time (s)Magnetic field (T)induced emf (V)
    -1010
    -510
    010
    11.030.0025434
    21.060.0025434
    31.090.0025434
    41.120.0025434
    51.150.0025434
    61.180.0025434
    71.210.0025434
    81.240.0025434
    91.270.0025434
    101.30.0025434
    151.30
    201.30

Formula used:

  A=πr2A=arear=radiusΦ=NBAN=number of turnsε=dΦdtε= emft= time

Calculation:

emf at 1 s as shown below, all the emf values were calculated like that by substituting the magnetic field at particular time.

  B at 1 s =1.03 TBy substituting,ε=dΦdtε=ddt(NBπr2)ε=Nπr2dBdtε=-30*3.14*(0.03 m)2(1.03 T1.00 T)1 sε=2.54 mV

Conclusion:

Graphs were drawn in the answer section.

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