(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

MENT OF EDI 3. A long, straight conductor carries a current of 8.0 A and is directed to the right. What is the magnitude of the magnetic field produced by the current-carrying wire 2.0 m from the wire? (See figure below) Draw the magnetic field to determine the direction of the field at the specified point P. RICHA AS SACY

Find the magnetic flux density when a point from a finite current length element of current 0.5A and radius 100 nm. 2.0 0.0 1.0 none of the above 0.5

5. A conductor 800 mm long moves across a uniform magnetic field with its axis making 35⁰ with the direction of the magnetic is 10,000 lines/in² and a voltage of 45 V appears across the conductor. Determine the velocity of the field. The flux density conductor in m/sec. 6. A periodic voltage has the following voltages: 15 V at 0.02 sec, 9 V at another 0.05 sec and 0 V for another 0.02 sec, the

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.