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 $\mathbf{\sigma }\mathbf{=}\mathbf{\infty }$and is J finite, then

$\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{(}\mathbf{v}\mathbf{\times }\mathbf{B}\mathbf{)}$

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

$\mathbf{d\Phi }\mathbf{=}{\mathbf{\int }}_{\mathbf{S}\mathbf{\text{'}}}\mathbf{B}\left(t+\mathrm{dt}\right)\mathbf{‐}\mathbf{da}\mathbf{-}{\mathbf{\int }}_{\mathbf{S}}\mathbf{B}\left(t\right)\mathbf{‐}\mathbf{da}$

Use $\mathbf{\nabla }\mathbf{·}\mathbf{B}\mathbf{=}\mathbf{0}$to show that

${\mathbf{\int }}_{\mathbf{S}\mathbf{\text{'}}}\mathbf{B}\left(t+\mathrm{dt}\right)\mathbf{‐}\mathbf{da}\mathbf{+}{\mathbf{\int }}_{\mathbf{R}}\mathbf{B}\left(t+\mathrm{dt}\right)\mathbf{‐}\mathbf{da}\mathbf{=}{\mathbf{\int }}_{\mathbf{S}}\mathbf{B}\left(t+\mathrm{dt}\right)\mathbf{‐}\mathbf{da}$

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

$\mathbf{d\Phi }\mathbf{=}\mathbf{dt}{\mathbf{\int }}_{\mathbf{S}}\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{t}}\mathbf{·}\mathbf{da}\mathbf{-}{\mathbf{\int }}_{\mathbf{R}}\mathbf{B}\left(t+\mathrm{dt}\right)\mathbf{‐}\mathbf{da}$

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

$\mathbf{dt}{\mathbf{\oint }}_{\mathbf{P}}\left(B\times v\right)\mathbf{·}\mathbf{dI}$

And invoke Stokes' theorem to conclude that

$\frac{\mathbf{d\Phi }}{\mathbf{dt}}\mathbf{=}{\mathbf{\int }}_{\mathbf{S}}\left(\frac{\partial B}{\partial t}-\nabla \times \left(v\times B\right)\right)\mathbf{·}\mathbf{da}$

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

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.

$\mathbf{J}\mathbf{=}\mathbf{\sigma }\left(E+v\times B\right)$ …… (1)

Here, $\mathit{\sigma }$ is charge density, Eis electrical field andB is magnetic field.

Write the formula of invoke stokes’ theorem.

$\mathbf{\int }\mathbf{B}\mathbf{·}\mathbf{da}\mathbf{=}\mathbf{0}$ …… (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 $\infty$for $\sigma$into equation (1).

$\left(\mathrm{E}+\mathrm{v}\times \mathrm{B}\right)=0$

Taking curl on both sides then

$\nabla \times \mathrm{E}+\nabla \times \left(\mathrm{v}\times \mathrm{B}\right)=0$

From faraday’s law

$\nabla \times E=-\frac{\partial B}{\partial t}$

Then:

$$\begin{gathered} -\frac{\partial B}{\partial t}+\nabla \times \left(\nabla \times B\right)=0 \\ \frac{\partial B}{\partial t}=\nabla \times \left(\nabla \times B\right) \end{gathered}$$

Therefore, the value to prove that $\frac{\partial B}{\partial t}=\nabla \times \left(\nabla \times B\right)$.

04

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

As we know that for any closed surface.

$\nabla \times \mathrm{B}=0$

Determine invoke stokes’ theorem.

Substitute ${\int }_{S\text{'}}B\left(t+dt\right)·da-{\int }_{R}B\left(t+dt\right)·da-{\int }_{S}B\left(t\right).da\mathrm{for}\int \mathrm{B}$into equation (2).

${\int }_{S\text{'}}B\left(t+dt\right)·da-{\int }_{R}B\left(t+dt\right)·da-{\int }_{S}B\left(t\right).da=0$

Here:

$$\begin{gathered} d\varphi ={\int }_{S\text{'}}B\left(t+dt\right)·da-{\int }_{S}B\left(t\right)·da-{\int }_{R}B\left(t+dt\right)·da \\ ={\int }_S\left[B\left(t+dt\right)·da-B\left(t\right)\right]·da-{\int }_{R}B\left(t+dt\right)·da \\ =\left\{{\int }_S\left(\frac{\partial B}{dt}·dt\right)dt-{\int }_{R}B\left(t+dt\right)·da\right\} \\ ={\int }_S\left(\frac{\partial B}{dt}·dt\right)dt-{\int }_{R}B\left(t+dt\right)·\left(\mathrm{dI}\times \mathrm{v}\right)dt \end{gathered}$$

Solve further as

Then:

$$\begin{gathered} d\varphi =dt{\int }_S\frac{\partial B}{\partial t}·da-dt{\int }_{S}B·\left(\mathrm{dI}\times \mathrm{v}\right) \\ d\varphi =dt\left\{{\int }_S\left(\frac{\partial B}{\partial t}\right)·da-{\int }_S\nabla \times \left(\mathrm{v}\times \mathrm{B}\right)·da\right\} \\ \frac{d\varphi }{dt}={\int }_S\left[\frac{\partial B}{\partial t}-\nabla \times \left(v\times B\right)\right]·da=0 \end{gathered}$$

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