Calculate the electric field intensity E ( r , θ , ϕ , t ) and the magnetic field intensity H ( r , θ , ϕ , t ) at the far field.

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Answer 1

Question

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

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Answer 2

Question

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

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Answer 3

Question

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

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Answer 4

Question

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

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Answer 5

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Answer 6

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Answer 7

Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

The electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

Given that,

$As=50e−jβrrax$ where $r2=x2+y2+z2$ .

Using vector transformation (in spherical coordinate system),

$ax=sinθcosϕar+cosθcosϕaθ−sinϕaϕ$

Therefore, the given vector function $As$ is written as,

$As=50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)$ (1)

Write the general expression for magnetic vector potential $As$ .

$∇×Asμ=Hs$ (2)

Here,

$μ$ is the permeability of the medium, and

$Hs$ is the magnetic field intensity.

Substitute equation (1) in (2).

$∇×(50e−jβrr(sinθcosϕar+cosθcosϕaθ−sinϕaϕ)As)μ=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕr(50e−jβrμrcosθcosϕ)rsinθ(−50e−jβrμrsinϕ)|=Hs1r2sinθ|arraθrsinθaϕ∂∂r∂∂θ∂∂ϕ50e−jβrμrsinθcosϕ50e−jβrμcosθcosϕ−50e−jβrμsinθsinϕ|=Hs1r2sinθ[(−50e−jβrμcosθsinϕ−50e−jβrμcosθsinϕ)ar−(j50βe−jβrμsinθsinϕ+50e−jβrμrsinθsinϕ)raθ+(−j50βe−jβrμcosθcosϕ−50e−jβrμrcosθcosϕ)rsinθaϕ]=Hs$

Reduce the equation as follows,

$−100cosθsinϕμr2sinθe−jβrar−50μr2(1−jβr)sinϕe−jβraθ−50μr2cosθcosϕ(1+jβr)e−jβraϕ=Hs$

At far field, $1r$ term only remains. Therefore,

$Hs=j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)$ (3)

Consider the general expression to calculate the electric fields intensity.

$Es=−ηar×Hs$ (4)

Here,

$η$ is the intrinsic impedance.

Substitute equation (3) in (4).

$Es=−ηar×[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)]$

$Es=−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)$ (5)

The general expression to calculate the electric field intensity is,

$E=Re[Esejωt]$ (6)

Substitute equation (5) in (6).

$E=Re[−j50βηe−jβrμr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re[−j50βηej(ωt−βr)μr(sinϕaϕ+cosθcosϕaθ)ejωt]=Re{−j50βη[cos(ωt−βr)+jsin(ωt−βr)]μr(sinϕaϕ+cosθcosϕaθ)ejωt}=50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$

The general expression to calculate the magnetic field intensity is,

$H=Re[Hsejωt]$ (7)

Substitute equation (3) in (7).

$H=Re[j50μrβe−jβr(sinϕaθ−cosθcosϕaϕ)ejωt]=Re[j50μrβej(ωt−βr)(sinϕaθ−cosθcosϕaϕ)]=Re{j50μrβ[cos(ωt−βr)+jsin(ωt−βr)](sinϕaθ−cosθcosϕaϕ)}=−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$

Conclusion:

Thus, the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field are $50βημrsin(ωt−βr)(sinϕaϕ+cosθcosϕaθ) V/m$ and $−50μrβsin(ωt−βr)(sinϕaθ−cosθcosϕaϕ) A/m$ respectively.

Answer 12

Ch. 13.5 - Prob. 1PE Ch. 13.5 - Prob. 2PE Ch. 13.6 - Prob. 3PE Ch. 13.6 - Prob. 4PE Ch. 13.6 - Prob. 5PE Ch. 13.7 - Prob. 6PE Ch. 13.8 - Prob. 8PE Ch. 13.8 - Prob. 9PE Ch. 13.9 - Prob. 10PE Ch. 13 - Prob. 1RQ

Calculate the electric field intensity E ( r , θ , ϕ , t ) and the magnetic field intensity H ( r , θ , ϕ , t ) at the far field.

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Calculate the electric field intensity $E(r, θ, ϕ, t)$ and the magnetic field intensity $H(r, θ, ϕ, t)$ at the far field.

Answer to Problem 1P

Explanation of Solution

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