Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
Question
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Chapter 7, Problem 7.31P
To determine

(a)

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

Expert Solution
Check Mark

Answer to Problem 7.31P

   ϕa=0.392μWb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1cm<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=1.66×105azA/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)ρA/m(102m<ρ<1.4×102m)

As B=μ0H

   B=0.104(ρ2 10 4)ρaϕWb/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 )21042104ln(1.21.0)]ϕa=3.92×107Wbϕa=0.392μWb

To determine

(b)

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

Expert Solution
Check Mark

Answer to Problem 7.31P

   ϕb=1.49μWb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1cm<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=1.66×105azA/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)ρA/m(102m<ρ<1.4×102m)

As B=μ0H

   B=0.104(ρ2 10 4)ρaϕWb/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 )21042104ln(1.41.0)]ϕb=1.49×106Wbϕb=1.49μWb

To determine

(c)

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

Expert Solution
Check Mark

Answer to Problem 7.31P

   ϕc=27μWb

Explanation of Solution

Given:

Total current carried by cylindrical shell is I=50A

Shell is defined by 1cm<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=1.66×105azA/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)ρA/m(102m<ρ<1.4×102m)

As B=μ0H

   B=0.104( ( 1.4× 10 2 ) 2 10 4 )ρaϕB= 10 5ρaϕWb/m2

Now,

   ϕc= B.dS= 0 1 1.4× 10 2 20× 10 2 10 5 ρ dρdz ϕc=105ln(201.4)ϕc=2.7×105Wbϕc=27μWb

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Chapter 7 Solutions

Engineering Electromagnetics

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