Initial magnetic field in the middle of the solenoid

Question

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Answer 1

Question

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

(f)

To determine

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

(f)

Expert Solution

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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Answer 2

Question

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

(f)

To determine

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

(f)

Expert Solution

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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Answer 3

Question

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

(f)

To determine

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

(f)

Expert Solution

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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Answer 4

Question

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

(f)

To determine

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

(f)

Expert Solution

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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Answer 5

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

(f)

To determine

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

(f)

Expert Solution

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

Answer 6

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

Answer 7

Chapter 20, Problem 19P

(a)

To determine

Initial magnetic field in the middle of the solenoid

Answer 8

(a)

Expert Solution

Answer to Problem 19P

Solution: Initial magnetic field in the middle of the solenoid is $3.77×10−3 T$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

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 $4π×10−7 T.m/A$ for $μo$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×2 A(20 cm)(10−2 m1 cm)=3.77×10−3 T$

Conclusion:

Initial magnetic field in the middle solenoid is $3.77×10−3 T$

Answer 13

(b)

To determine

Final magnetic field in the solenoid

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

Answer 14

(b)

To determine

Final magnetic field in the solenoid

Answer 15

(b)

Expert Solution

Answer to Problem 19P

Solution: Final magnetic field in the solenoid is $9.42×10−3 T$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

Formula for final magnetic field in the solenoid,

$Bf=NμoIl$

Substitute 5A for I, , 300 for N, 20cm for l and $4π×10−7 T.m/A$ for $μ$ in the above expression to get $Bi$

$Bi=4π×10−7 T.m/A×300×5 A(20 cm)(10−2 m1 cm)=9.42×10−3 T$

Conclusion:

Final magnetic field in the solenoid is $9.42×10−3 T$

Answer 20

(c)

To determine

Area of 4-turn coil

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

Answer 21

(c)

To determine

Area of 4-turn coil

Answer 22

(c)

Expert Solution

Answer to Problem 19P

Solution: Area of 4-turn coil is $7.07×10−4 m2$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given Info: radius of coil is $1.50×10−2 m$

Formula for area of 4-turn coil,

$A=πr2$

Substitute $1.50×10−2 m$ for r in the above expression to get A.

$A=π(1.50×10−2 m)2=7.07×10−4 m2$

Conclusion:

Area of 4-turn coil is $7.07×10−4 m2$

Answer 27

(d)

To determine

Change in magnetic flux through 4- turn coil.

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

Answer 28

(d)

To determine

Change in magnetic flux through 4- turn coil.

Answer 29

(d)

Expert Solution

Answer to Problem 19P

Solution: Change in magnetic flux through 4- turn coil is $3.99×10−6 Wb$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given Info: initial magnetic field is $3.77×10−3 T$ , final magnetic field is $9.42×10−3 T$ and area of coil is $7.07×10−4 m2$

Formula for change in magnetic flux,

$Δϕ=ΔBA$

Simplify above expression

$Δϕ=(Bf−Bi)A$

Substitute $9.42×10−3 T$ for $Bf$ , $3.77×10−3 T$ for $Bi$ and $7.07×10−4 m2$ for A in the above expression to get $Δϕ$

$Δϕ=(9.42×10−3 T−3.77×10−3 T)7.07×10−4 m2=3.99×10−6 Wb$

Conclusion:

Change in magnetic flux through each turn of coil is $3.99×10−6 Wb$

Answer 34

(e)

To determine

Average Induced emf in 4-turn coil.

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

Answer 35

(e)

To determine

Average Induced emf in 4-turn coil.

Answer 36

(e)

Expert Solution

Answer to Problem 19P

Solution: Average Induced emf in 4-turn coil is $1.77×10−5 V$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given Info: magnetic flux is $3.99×10−6 Wb$

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 $3.99×10−6 Wb$ for $Δϕ$ , 4 for N , and $0.900 s$ for $Δt$ in the above expression to get $ε$

$ε=4(3.99×10−6 Wb0.900 s)=1.77×10−5 V$

Conclusion:

Average Induced emf in 4-turn coil is $1.77×10−5 V$

Answer 41

(f)

To determine

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

(f)

Expert Solution

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

Answer 42

(f)

To determine

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

Answer 43

(f)

Expert Solution

Answer to Problem 19P

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

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

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