Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 7, Problem 41P

(a)

To determine

Find whether the given field D is electrostatic or magnetostatic field in free space.

(a)

Expert Solution
Check Mark

Answer to Problem 41P

The vector D is possibly a magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

D=y2zax+2(x+1)yzax(x+1)z2az

Generally, the vector is a magnetostatic field if its dot product is zero and the vector is an electrostatic field if its cross product is zero. That is,

For magnetostatic field, (vector)=0

For electrostatic field, ×(vector)=0

Substitute y2zax+2(x+1)yzax(x+1)z2az for D to find D.

D=xDx+yDy+zDz=x(y2z)+y(2(x+1))+z((x+1))=0+0+0=0

Substitute y2zax+2(x+1)yzax(x+1)z2az for D to find ×D.

×D=|axayazxyzDxDyDz|=|axayazxyzy2z2(x+1)yz(x+1)z2|=[(y((x+1)z2)z(2(x+1)yz))ax(x((x+1)z2)z(y2z))ay+(x(2(x+1)yz)y(y2z))az]=[((0)2(x+1)y(1))ax(((1)z2+0)y2(1))ay+((2(1)yz+0)(2y)z)az]

Simplify the above equation.

×D=(2(x+1)y)ax(z2y2)ay+(2yz2yz)az=(2(x+1)y)ax+(z2+y2)ay+(0)az=2(x+1)yax+(z2+y2)ay0

Therefore, for the field D,

D=0×D0

Conclusion:

Thus, the vector D is possibly a magnetostatic field.

(b)

To determine

Find whether the given field E is electrostatic or magnetostatic field in free space.

(b)

Expert Solution
Check Mark

Answer to Problem 41P

The vector E is possibly a magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

E=(z+1)ρcosϕaρ+sinϕρaz

Substitute (z+1)ρcosϕaρ+sinϕρaz for E to find E.

E=1ρρ(ρEρ)+1ρϕEϕ+zEz=1ρρ(ρ((z+1)ρcosϕ))+1ρϕ(0)+z(sinϕρ)=1ρρ((z+1)cosϕ)+0+0=0

Substitute (z+1)ρcosϕaρ+sinϕρaz for E to find ×E.

×E=1ρ|aρρaϕazρϕz(z+1)ρcosϕρ(0)sinϕρ|=1ρ|aρρaϕazρϕz(z+1)ρcosϕ0sinϕρ|=1ρ[(ϕ(sinϕρ)z(0))aρ(ρ(sinϕρ)z((z+1)ρcosϕ))ρaϕ+(ρ(0)ϕ((z+1)ρcosϕ))az]=1ρ[ϕ(sinϕρ)aρ(ρ(sinϕρ)z((z+1)ρcosϕ))ρaϕϕ((z+1)ρcosϕ)az]

Simplify the above equation.

×E=1ρ[cosϕρaρρ(sinϕ(1ρ2)(cosϕρ)(1+0))aϕ(z+1)ρ(sinϕ)az]=1ρ[cosϕρaρρ(sinϕρ2(cosϕρ))aϕ+(z+1)sinϕρaz]=cosϕρ2aρ(sinϕρ2(cosϕρ))aϕ+(z+1)sinϕρ2az0

Therefore, for the field E,

E=0×E0

Conclusion:

Thus, the vector E is possibly a magnetostatic field.

(c)

To determine

Find whether the given field F is electrostatic or magnetostatic field in free space.

(c)

Expert Solution
Check Mark

Answer to Problem 41P

The vector F is neither electrostatic nor magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

F=1r2(2cosθar+sinθaθ)=2r2cosθar+1r2sinθaθ

Substitute 2r2cosθar+1r2sinθaθ for F to find F.

F=1r2r(r2Fr)+1rsinθθ(sinθFθ)+1rsinθϕ(Fϕ)=1r2r(r2(2r2cosθ))+1rsinθθ(sinθ(1r2sinθ))+1rsinθϕ(0)=1r2r(2cosθ)+1rsinθθ(1r2sin2θ)+0=1r2(0)+1r3sinθ(2sinθcosθ)

Simplify the above equation.

F=0+1r3(2cosθ)=2cosθr30

Substitute 2r2cosθar+1r2sinθaθ for F to find ×F.

×F=1r2sinθ|arraθrsinθaϕrθϕ2r2cosθr(1r2sinθ)0|=1r2sinθ|arraθrsinθaϕrθϕ2r2cosθ1rsinθ0|=1r2sinθ[(θ(0)ϕ(1rsinθ))ar(r(0)ϕ(2r2cosθ))raθ+(r(1rsinθ)θ(2r2cosθ))rsinθaϕ]=1r2sinθ[(0(0))ar(0(0))raθ+(sinθ(1r2)(2r2)(sinθ))rsinθaϕ]

Simplify the above equation.

×F=1r2sinθ[0+0+rsinθ(1r2sinθ+2r2sinθ)aϕ]=rsinθr2sinθ(1r2sinθ+2r2sinθ)aϕ=1r3(sinθ)aϕ0

Therefore, for the field F,

F0×F0

Conclusion:

Thus, the vector F is neither electrostatic nor magnetostatic field.

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