College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Question
Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution
Check Mark

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is 3.77×103T

Explanation of Solution

Given Info: current in the coil is 2A, number of turns is 300, length of the coil is 20cm .

Formula for initial magnetic field in the solenoid,

Bi=NμoIl

  • l is the length of the coil
  • N is the number of turns in the coil
  • I is the current in the coil

Substitute 2A for I, 300 for N, 20cm for l and 4π×107T.m/A for μo in the above expression to get Bi

Bi=4π×107T.m/A×300×2A(20cm)(102m1cm)=3.77×103T

Conclusion:

Initial magnetic field in the middle solenoid is 3.77×103T

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution
Check Mark

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is 9.42×103T

Explanation of Solution

Given Info: current in the coil is 5A, number of turns is 300, length of the coil is 20cm

Formula for final magnetic field in the solenoid,

Bf=NμoIl

Substitute 5A for I, , 300 for N, 20cm for l and 4π×107T.m/A for μ in the above expression to get Bi

Bi=4π×107T.m/A×300×5A(20cm)(102m1cm)=9.42×103T

Conclusion:

Final magnetic field in the solenoid is 9.42×103T

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution
Check Mark

Answer to Problem 19P

Solution: Area of 4-turn coil is 7.07×104m2

Explanation of Solution

Given Info: radius of coil is 1.50×102m

Formula for area of 4-turn coil,

A=πr2

Substitute 1.50×102m for r in the above expression to get A.

A=π(1.50×102m)2=7.07×104m2

Conclusion:

Area of 4-turn coil is 7.07×104m2

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution
Check Mark

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is 3.99×106Wb

Explanation of Solution

Given Info: initial magnetic field is 3.77×103T , final magnetic field is 9.42×103T and area of coil is  7.07×104m2

Formula for change in magnetic flux,

Δϕ=ΔBA

Simplify above expression

Δϕ=(BfBi)A

Substitute 9.42×103T for Bf , 3.77×103T for Bi and 7.07×104m2 for A in the above expression to get Δϕ

Δϕ=(9.42×103T3.77×103T)7.07×104m2=3.99×106Wb

Conclusion:

Change in magnetic flux through each turn of coil is 3.99×106Wb

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution
Check Mark

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is 1.77×105V

Explanation of Solution

Given Info: magnetic flux is  3.99×106Wb

Formula for Average Induced emf in 4-turn coil

ε=N(ΔϕBΔt)

  • ΔϕB is the change in magnetic flux
  • Δt is the change in time

Substitute 3.99×106Wb for Δϕ , 4 for N , and 0.900s for Δt in the above expression to get ε

ε=4(3.99×106Wb0.900s)=1.77×105V

Conclusion:

Average Induced emf in 4-turn coil is 1.77×105V

(f)

To determine

Why the contribution of the magnetic field by the current in the 4-turn coil be neglected.

(f)

Expert Solution
Check Mark

Answer to Problem 19P

Contributions to the magnetic field by the current in the 4-turn coil is negligible

Explanation of Solution

The magnitude of the magnetic field depends on the amount of induced current in the coil. From above calculations, the emf induced in the coil is small according to part (e). When the emf is small, the current in the 4-turn will also be small. Hence, the magnetic field generated by the current is negligible.

Conclusion:

Contributions to the magnetic field by the current in the 4-turn coil is negligible

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Chapter 20 Solutions

College Physics

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