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

Question

Want to see more full solutions like this?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

For question 1B Draw paraboloid Z=x^2+y^2

A3 ϕ, 15 kV, 60 Hz, 30 MVA, Y-connected cylindrical-rotor synchronous machine generator has Ra =0.5 Ω per phase and Xs =5.2 Ω per phase. The generator delivers the rated load at 15 kV and 0.85 lagging power factor. Determine the excitation voltage, Ef, for this operating condition.

A3 φ, 50 Hz infinite bus bar is connected to two synchronous generators. Generator 1 has a speed of 300 rpm and generator 2 has 30 poles. Determine: the speed of generator 2 the number of poles of generator 1

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

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.