Prove Alfven's theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, "frozen" into the fluid.)

(a) Use Ohm's law, in the form of Eq. 7.2, together with Faraday's law, to prove that if σ=and is J finite, then

Bt=×(v×B)

(b) Let S be the surface bounded by the loop (P)at time t , and S'a surface bounded by the loop in its new position (P')at time t+dt (see Fig. 7.58). The change in flux is

=S'B(t+dt)da-SB(t)da

Use ·B=0to show that

S'B(t+dt)da+RB(t+dt)da=SB(t+dt)da

(Where R is the "ribbon" joining P and P' ), and hence that

=dtSBt·da-RB(t+dt)da

(For infinitesimal dt ). Use the method of Sect. 7.1.3 to rewrite the second integral as

dtP(B×v)·dI

And invoke Stokes' theorem to conclude that

dt=S(Bt-×v×B)·da

Together with the result in (a), this proves the theorem.

Short Answer

Expert verified

(a) The value to prove that Bt=×(v×B).

(b) The value of invoke stokes’ theorem is dϕdt=SBt-×v×B·da=0.

Step by step solution

01

Write the given data from the question.

Let S be the surface bounded by the loop (P) at time t .

Let S' a surface bounded by the loop in its new position (P') at time t+dt .

Let R is the "ribbon" joining P and P' .

02

Determine the formula of ohm’s law using faraday’s law and formula of invoke stokes’ theorem.

Write the formula ofohm’s law using faraday’s law.

J=σ(E+v×B) …… (1)

Here, σ is charge density, Eis electrical field andB is magnetic field.

Write the formula of invoke stokes’ theorem.

B·da=0 …… (2)

Here,B is magnetic field and da is the radius of the circle.

03

(a) Determine the value to prove that ∂B∂t=∇×(∇×B) .

According to ohm’s law

Substitute 0 for J and for σinto equation (1).

E+v×B=0

Taking curl on both sides then

×E+×v×B=0

From faraday’s law

×E=-Bt

Then:

-Bt+××B=0Bt=××B

Therefore, the value to prove that Bt=××B.

04

(b) Determine the value of invoke stokes’ theorem.

As we know that for any closed surface.

×B=0

Determine invoke stokes’ theorem.

Substitute S'Bt+dt·da-RBt+dt·da-SBt.daforBinto equation (2).

S'Bt+dt·da-RBt+dt·da-SBt.da=0

Here:

dϕ=S'Bt+dt·da-SBt·da-RBt+dt·da=SBt+dt·da-Bt·da-RBt+dt·da=SBdt·dtdt-RBt+dt·da=SBdt·dtdt-RBt+dt·dI×vdt

Solve further as

Then:

dϕ=dtSBt·da-dtSB·dI×vdϕ=dtSBt·da-S×v×B·dadϕdt=SBt-×v×B·da=0

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Most popular questions from this chapter

A perfectly conducting spherical shell of radius rotates about the z axis with angular velocity ω, in a uniform magnetic field B=B0Z^. Calculate the emf developed between the “north pole” and the equator. Answer:localid="1658295408106" [12B0ωα2].

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(b) Find the inductance per unit length, L .

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A toroidal coil has a rectangular cross section, with inner radius a , outer radius a+w, and height h . It carries a total of N tightly wound turns, and the current is increasing at a constant rate (dl/dt=k). If w and h are both much less than a , find the electric field at a point z above the center of the toroid. [Hint: Exploit the analogy between Faraday fields and magnetostatic fields, and refer to Ex. 5.6.]

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An alternating current I(t)=I0cos(ωt) (amplitude 0.5 A, frequency ) flows down a straight wire, which runs along the axis of a toroidal coil with rectangular cross section (inner radius 1cm , outer radius 2 cm , height 1 cm, 1000 turns). The coil is connected to a 500Ω resistor.

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