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13—Vector Calculus 2 343

13.24 The same field and surface as the preceding problem, but now the surface integraldA~×F~.


Ans:zˆ 2 πBr^3 / 3


13.25 (a) Prove the identity∇.


(~


A×B~


)

=B~.∇×A~−A~.∇×B~. (index mechanics?)


(b) Next apply Gauss’s theorem to∇.


(~


A×B~


)

and take the special case thatB~is an arbitrary constant


to derive Eq. (13.19).


13.26 (a) Prove the identity∇.(fF~) =f∇.F~+F~.∇f.


(b) Apply Gauss’s theorem to∇.(fF~)for an arbitrary constantF~to derive a result found in another


problem.
(c) Explain why the word “arbitrary” is necessary here.


13.27 The vector potential is not unique, as you can add an arbitrary gradient to it without affecting


its curl. Suppose thatB~=∇×A~with


A~=xαxyzˆ +ˆyβx^2 z+zγxyzˆ^2


Find a functionf(x,y,z)such thatA~′=A~+∇fhas thez-component identically zero. Do you get


the sameB~ by taking the curl ofA~and ofA~′?


13.28 Take the vector field


B~=αxyxˆ+βxyˆy+γ(xz+yz)ˆz


Write out the equationB~=∇×A~in rectangular components and figure out what functionsAx(x,y,z),


Ay(x,y,z), andAz(x,y,z)will work. Note: From the preceding problem you see that you may if you


wish pick any one of the components ofA~to be zero — that will cut down on the labor. Also, you


should expect that this problem is impossible unlessB~has zero divergence. That fact should comeout


of your calculations, so don’t put it in yet. Determine the conditions onα,β, andγthat make this


problem solvable, and show that this is equivalent to∇.B~= 0.


13.29 A magnetic monopole, if it exists, will have a magnetic fieldμ 0 qmr/ˆ 4 πr^2. The divergence of this


magnetic field is zero except at the origin, but that means that not every closed surface can be shrunk
to a point without running into the singularity. The necessary condition for having a vector potential is
not satisfied. Try to construct such a potential anyway. Assume a solution in spherical coordinates of


the formA~=φfˆ (r)g(θ)and figure out whatfandgwill have thisB~for a curl. Sketch the resulting


A~. You will run into a singularity (or two, depending). Ans:A~=φμˆ 0 qm(1−cosθ)/( 4 πr^2 sinθ) (not


unique)


13.30 Apply the Reynolds transport theorem to the other of Maxwell’s equations.


∇×B~=μ 0 ~j+μ 0  0


∂E~


∂t


Don’t simply leave the result in the first form that you find. Manipulate it into what seems to be the


best form. Useμ 0  0 = 1/c^2.


Ans:


∮(~


B−~v×E/c~^2


)

.d~`=μ 0 ∫


(

~j−ρ~v


)

.dA~+μ 0  0 (d/dt)∫E~.dA~

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