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For Theorem 2, show thatd⇒a,a⇒c,c⇒b,b⇒candc⇒a

Short Answer

Expert verified

The statement a⇒chas been shown. The statements c⇒b and b⇒chas been shown. The statement (c)⇒(a) has been shown.

Step by step solution

01

Describe the given information.

The theorem (2), describing the line integral of a vector along a closed path, shows that for a closed path d⇒a, a⇒c, c⇒b, b⇒cand c⇒a.

02

Define the solenoidal field.

The solenoid fields or divergence less fields are those which have zero divergence, that is ∇⋅F=0 . If divergence theorem is applied, then surface integral of the function is 0, that is, ∫F⋅da=0. It can be expressed as a curl of another vectorF=∇×A.

03

Prove d⇒a.

As F is expressed as curl of some vector A , that is, F=∇×Aand divergence of F is 0, that is∇⋅F=0, find the curl of function F as,

Substitute ∇×A for F into ∇.F=0

∇.(∇×A)=0

The result is true because the divergence of curl of a vector is always zero. Thus we can say thatd⇒a.

According to gauss divergence theorem,

∭∇⋅FdV=∫F⋅da

Since the divergence of vector function F is zero, that is∇⋅F=0.

Substitute 0 for ∇.Finto ∭∇⋅FdV=∫F⋅da.

∭0dV=∫F⋅da∫F⋅da=0

Thus, we can say a⇒c.

04

Prove (c)⇒(b) and (b)⇒(c).

As discussed, the surface integral vector F is 0. So∫cF⋅da=0 . If the defined surface, have inward surface vector at front surface (1) and outward surface vector at back surface (2). Then, the total surface vector can be expressed as a sum of surface vectors of surface (1) and (2), as,

∫cF⋅da=∫(1)F⋅da−∫(2)F⋅da

Substitute 0 for∫cF⋅da into∫cF⋅da=∫(1)F⋅da−∫(2)F⋅da .

0=∫(1)F⋅da−∫(2)F⋅da∫(1)F⋅da=∫(2)F⋅da

As areal vector at both the surface is 0, path from (1) to (2) is independent of surface. Hence, from c⇒b.

As the line integral ∫cF⋅da is independent of path from (1) to (2) then for any closed loop, this line integral gives same value as the final and initial points in any closed loop coincide with each other.

Thus, we can write b⇒c.

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Most popular questions from this chapter

Question: Evaluate the following integrals:

(a)∫-22(2x+3)δ(3x)dx

(b)∫02(x3+3x+2)δ(1-x)dx

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