Question 4.1.1: A straight cylindrical magnetic flux tube with uniform axial......
A straight cylindrical magnetic flux tube with uniform axial magnetic field, B_{z} = B_{0}, is embedded into a perfectly conducting fluid.
a) The tube is twisted around its axis by a twist angle \phi(z) as shown in Figure 4.1. What is the deformed magnetic field?
b) After such, the tube undergoes uniform radial expansion from r → 2r. Derive the resulting magnetic field.
a) Consider two neighboring fluid particles, A and B, that are initially separated by a distance dz along the field (see Figure 4.1). By using cylindrical coordinates (r, θ, z), their separation before and after the twisting deformation is
δ\vec{l}^{(0)} = (0, 0, dz); δ\vec{l}^{(a)} = (0, r_{0} \frac{d\phi}{dz} dz, dz)
Since such a deformation is incompressible, it follows from Equation (4.4) that
B_{i} = \frac{ρ}{ρ_{0}} B^{(0)}_{k} \frac{δl_{i}}{δl_{k}^{(0)}} , (4.4)
B^{(a)}_{z} = B_{0} \frac{δl^{(a)}_{z}}{δl^{(0)}_{z}} = B_{0}; B^{(a)}_{θ} = B_{0} \frac{δl^{(a)}_{θ}}{δl^{(0)}_{z}}= B_{0} r_{0} \frac{d\phi}{dz}
b) Radial expansion from r → 2r leads to the separation vector δ\vec{l}^{(b)} = (0, 2r_{0} \frac{d\phi}{dz} dz, dz) and a density reduction ρ = ρ_{0}/4. Therefore, by applying the transformation rule (4.4) once again, one gets:
B^{(b)}_{z} = \frac{B_{0}}{4}; B^{(b)}_{θ} = \frac{B^{(a)}_{θ}}{4} \frac{δl^{(b)}_{θ}}{δl^{(a)}_{θ}} = \frac{B_{0}}{2} r_{0} \frac{d\phi}{dz} = \frac{B_{0}}{4} r \frac{d\phi}{dz}