(a) The total magnetic flux crossing plane ϕ = 0 , 0 < z < 1 .

Question

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Answer 1

Question

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

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Students have asked these similar questions

please see fig 1 to solve

A particular battery charger produces a constant current of 2.4 amperes to charge a 3.7 volt Iphone battery. It can fully charge a dead battery to full charge in 6 hours. (a) How many electrons are in the charged battery? (b) What amount of energy in Joules does the battery provide if it deliveres 0.25 amperes of current to a phone for 4 hours.

please see fig 2 to answer

Answer 2

Question

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

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Answer 3

Question

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

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Answer 4

Question

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

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Answer 5

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

Answer 6

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

Answer 7

Chapter 7, Problem 7.31P

To determine

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer 8

Expert Solution

Answer to Problem 7.31P

ϕa=0.392μ Wb

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

0<ρ<1.2cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕa=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.2× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕa=0.104[( 1.2× 10 −2 )2−10−42−10−4ln(1.21.0)]ϕa=3.92×10−7 Wbϕa=0.392μ Wb

Answer 13

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

Answer 14

To determine

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer 15

Expert Solution

Answer to Problem 7.31P

ϕb=1.49μ Wb

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1<ρ<1.4cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104(ρ2− 10 −4)ρaϕ Wb/m2

Now,

ϕb=∫ ∫ B.dS= ∫ 0 1 ∫ 10 −2 1.4× 10 −2 0.104[ ρ− 10 −4 ρ ]dρdz ϕb=0.104[( 1.4× 10 −2 )2−10−42−10−4ln(1.41.0)]ϕb=1.49×10−6 Wbϕb=1.49μ Wb

Answer 20

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

Answer 21

To determine

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer 22

Expert Solution

Answer to Problem 7.31P

ϕc=27μ Wb

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1 cm < ρ <1.4 cm.

The plane is ϕ=0, 0<z<1.

1.4cm<ρ<20cm

Calculation:

The current density can be calculated as

J=Iπr2az

J=50π[ ( 1.4× 10 −2 ) 2− ( 1.0× 10 −2 ) 2]azJ=1.66×105az A/m2

The value of current density is 0 when the radii is less than 1 cm.

Applying Ampere's Circuital law, the value of Hϕ at the give radius will be

2πρHϕ=Iencl=∫02π ∫ 10 −2 ρ 1.66× 10 5 ρ ′ d ρ ′ dϕHϕ=8.30×104(ρ2− 10 −4)ρ A/m (10−2m<ρ<1.4×10−2m)

As B=μ0H

B=0.104( ( 1.4× 10 −2 ) 2 − 10 −4 )ρaϕ B= 10 −5ρaϕ Wb/m2

Now,

ϕc=∫ ∫ B.dS= ∫ 0 1 ∫ 1.4× 10 −2 20× 10 −2 10 −5 ρ dρdz ϕc=10−5ln(201.4)ϕc=2.7×10−5 Wbϕc=27μ Wb

Answer 27

Ch. 7 - Find H in rectangular components at P(2,3,4) if...Ch. 7 - Prob. 7.2P Ch. 7 - Prob. 7.3P Ch. 7 - Prob. 7.4P Ch. 7 - The parallel filamentary conductors shown in...Ch. 7 - A disk of radius a lies in the xy plane, with z...Ch. 7 - Prob. 7.7P Ch. 7 - For the finite-length current element on the z...Ch. 7 - Prob. 7.9P Ch. 7 - Prob. 7.10P

(a) The total magnetic flux crossing plane ϕ = 0 , 0 &lt; z &lt; 1 .

(a) The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer to Problem 7.31P

Explanation of Solution

(b) The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer to Problem 7.31P

Explanation of Solution

(c) The total magnetic flux crossing plane ϕ=0, 0<z<1.

Answer to Problem 7.31P

Explanation of Solution

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Chapter 7 Solutions

(a) The total magnetic flux crossing plane ϕ = 0 , 0 < z < 1 .

(a)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

(b)

The total magnetic flux crossing plane ϕ=0, 0<z<1.

(c)

The total magnetic flux crossing plane ϕ=0, 0<z<1.