# Frozen in Magnetic Field Lines

A perfectly conducting fluid undergoes an axisymmetric motion and contains an azimuthal magnetic field $$\textbf{B}_θ$$ . Show that $$\dfrac{B_θ}{r}$$ is conserved by each fluid element.

Question from Introduction to Magnetohydrodynamics by P.A Davidson.

Since it is perfectly conducting, we know from Alfvén's theorem flux should be conserved,since magnetic field lines and the direction of the fluid is in the same direction.

$$\int_{S}^{} \textbf B d \textbf S= constant$$

However, this information becomes useless, since magnetic field direction and direction of the surface are perpendicular to each other, so I can get no where and there is no other information available. What is the available given information to solve the problem? What am I missing?

Note
In cylindrical coordinates, the curl with only azimuthal field goes to: $$\nabla \times \mathbf{B} = - \hat{\mathbf{r}} \frac{ \partial B_{\theta} }{ \partial z } + \hat{\mathbf{z}} \frac{ 1 }{ r } \frac{ \partial }{ \partial r } \left( r B_{\theta} \right) \tag{1a}$$

Since we can assume that the field will not vary along $$\hat{\mathbf{z}}$$, the only term remaining is the second. If $$B_{\theta}$$ is constant with respect to radial distance, i.e., no local sources, then Equation 1a reduces to: $$\nabla \times \mathbf{B} = \hat{\mathbf{z}} \frac{ B_{\theta} }{ r } \tag{1b}$$

Flux Freezing
The time variation of the flux can be written as: \begin{align} \frac{ \partial \psi }{ \partial t_{1} } & = \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \frac{ \partial \mathbf{B} }{ \partial t } \tag{2a} \\ & = - \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \left( \nabla \times \mathbf{E} \right) \tag{2b} \end{align} where $$\hat{\mathbf{n}}$$ is the outward unit normal of the surface $$S$$.

Suppose our surface $$S$$ was defined by a closed contour, $$C$$, with infinitesimal length, $$d\mathbf{l}$$. If this contour moves along with an arbitrary fluid element at velocity $$\mathbf{V}$$, then the area swept out per unit time by $$C$$ would be given by $$\mathbf{V} \times d\mathbf{l}$$. Therefore, we can write another part of the flux given by: $$\frac{ \partial \psi }{ \partial t_{2} } = \oint_{C} \ \mathbf{B} \cdot \left( \mathbf{V} \times d\mathbf{l} \right) = \oint_{C} \ \left( \mathbf{B} \times \mathbf{V} \right) \cdot d\mathbf{l} \tag{3}$$ Note that the $$\partial_{t}$$ was pulled in to define the $$\mathbf{V}$$ from the area here.

Using Stokes' theorem we can rewrite Equation 3 as: $$\frac{ \partial \psi }{ \partial t_{2} } = \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \nabla \times \left( \mathbf{B} \times \mathbf{V} \right) \tag{4}$$

Then if we combine Equations 2b and 4 we have the total time rate of change of the flux, given by: $$\frac{ d \psi }{ dt } = - \int_{S} \ dA \ \hat{\mathbf{n}} \cdot \left[ \nabla \times \left( \mathbf{E} + \mathbf{V} \times \mathbf{B} \right) \right] \tag{5}$$

The frozen-in condition requires that the following holds: $$\mathbf{E} + \mathbf{V} \times \mathbf{B} = 0 \tag{6}$$ for a constant flux. If $$\mathbf{V}$$ is parallel to $$\mathbf{B}$$ then $$\mathbf{E} = 0$$ and the curl of $$\mathbf{V} \times \mathbf{B}$$ goes to (using standard vector calculus identities): $$\nabla \times \left( \mathbf{V} \times \mathbf{B} \right) = - \left( \mathbf{V} \cdot \nabla \right) \mathbf{B} \tag{7}$$ because all other terms go to zero from geometry/symmetry and Maxwell's equations. The $$\theta$$ term of the gradient is given by: $$\left( \nabla f \right)_{\theta} = \frac{ 1 }{ r } \frac{ \partial f }{ \partial \theta } \tag{8}$$

The rest is just symbol gymnastics to show that $$\tfrac{B_{\theta}}{r}$$ is a constant.