Question

Consider a cylindrical wire of radius R (indefinitely long) that carries a total steady current I such that there is a constant current density j across the profile of the wire (for the first part of this task, consider just a current density in vacuum)

a) in order to calculate the magnetic induction it is suitable to work in cylindrical coordinates. Considering Boundary conditions at ρ→∞, the magnetic induction ca be written as B=B_ρ (ρ,φ,z) e_ ρ + B_ φ(ρ,φ,z)e_ φ

Use the symmetry of the infinitely long cylindrical wire(and the corresponding current density) to simplify this ansatz for the magnetic induction.

b) use maxwell´s equation (no magnetic monopoles) to show that there is no component of the magnetic induction in the radial direction.

c) How does the result of your calculation change if only the conductor of the wire has a magnetic susceptiblity χ>0? Hint: use the material equations and the equation for linear media.

Answer 1

a) The given ansatz is

B = B(4,2) ep + B (2,6,2) ep (1)

First we consider the cylindrical symmetry, which implies that every value of the angular variable is equivalent. If we observe the wire from a position with coordinates
( , ,م)
and then move in the angular direction so that our coordinates become
(4+0,2)
, then because of cylindrical symmetry the physical observables will not change, hence the magnetic field will be independent of $\varphi$ , i.e.

B = B (2, 2) ep + B (P.2) ep (2)

The second symmetry in the system comes from the fact that the wire is indefinitely long. Hence the z-coordinates $z \text{ and } z+z_{1}$ are equivalent as far as physical observables are concerned, therefore

$\mathbf{B} = B_{\rho}\left(\rho \right )\mathbf{e}_{\rho}+B_{\varphi}\left(\rho \right )\mathbf{e}_{\varphi} \quad (3)$

b) The Maxwell's equations are

V. B = 0 = 1 двор) dB 10 (рвер)) р др + 0: Вpp) р + 0 е де dp

$\Rightarrow B_{\rho}(\rho)d\rho+\rho dB_{\rho}(\rho) = 0\Rightarrow d(B_{\rho}(\rho)\rho) = 0\Rightarrow B_{\rho}(\rho) = \frac{A}{\rho} \quad (4)$

is a constant of integration. We can see that as $\rho\to 0$ the solution in equation (4) goes to infinity, since that is unphysical, we set , hence

$B_{\rho}(\rho) = 0 \quad (5)$

The other Maxwell equaton is

$\nabla\times\mathbf{B} = \mathbf{j} \quad (6)$

We will use its integral form

$\oint_{C}\mathbf{B}\cdot d\mathbf{l} = \mu_{0}I_{enc} \quad (6a)$

is an Amperian circular loop of radius $\rho$ and concentric with the cylindrical wire. We can have two cases

i) $\rho< R$ : Equation (6a) becomes

μορ Βρ(ρ)2πρ = μο- πρ? Βρ(ρ) - (7) πR2 27R2

ii) $\rho> R$ : Equation (6a) becomes

$B_{\varphi}(\rho)2\pi \rho = \mu_{0}I\Rightarrow B_{\varphi}(\rho) = \frac{\mu_{0}I}{2\pi \rho} \quad (8)$

Hence

$\mathbf{B} = \left\{\begin{matrix} \frac{\mu_{0}I \rho}{2\pi R^{2}}\mathbf{e}_{\varphi},\; \rho\leq R\\ \\ \frac{\mu_{0}I }{2\pi \rho}\mathbf{e}_{\varphi},\; \rho > R \end{matrix}\right. \quad (9)$

c) In a linear medium with susceptiblity $\chi$ , the Maxwell equations become

$\nabla\cdot\mathbf{B} = 0 \quad (10)$

$\nabla\times\mathbf{H} = \mathbf{j}_{free} \quad (11)$

$\mathbf{B} = \mu_{0}\left(1+\chi \right )\mathbf{H} \quad (12)$

From symmetry of the problem and and the no monopole Maxwell equation we get

$\mathbf{B} = B_{\varphi}(\rho)\mathbf{e}_{\varphi}\Rightarrow \mathbf{H} = H_{\varphi}(\rho)\mathbf{e}_{\varphi} \quad (13)$

just like the previous case. Now we use the integral form of equation (11), to get

$\oint_{C}\mathbf{H}\cdot d\mathbf{l} = I_{enc.,free} \quad (14)$

where is an Amperian circular loop of radius $\rho$ and concentric with the cylindrical wire. Again we can have two cases

i) $\rho< R$ : Equation (14) becomes

$H_{\varphi}(\rho)2\pi \rho = \frac{I}{\pi R^{2}}\pi \rho^{2}\Rightarrow H_{\varphi}(\rho) = \frac{I\rho}{2\pi R^{2}} \quad (15)$

ii) $\rho> R$ : Equation (14) becomes

$H_{\varphi}(\rho)2\pi \rho = I\Rightarrow H_{\varphi}(\rho) = \frac{I}{2\pi \rho} \quad (16)$

Hence

$\mathbf{H} = \left\{\begin{matrix} \frac{I \rho}{2\pi R^{2}}\mathbf{e}_{\varphi},\; \rho< R\\ \\ \frac{I }{2\pi \rho}\mathbf{e}_{\varphi},\; \rho > R \end{matrix}\right. \quad (17)$

From equation (12), we get

$\mathbf{B} = \left\{\begin{matrix} \frac{\mu_{0}(1+\chi)I \rho}{2\pi R^{2}}\mathbf{e}_{\varphi},\; \rho< R\\ \\ \frac{\mu_{0}I }{2\pi \rho}\mathbf{e}_{\varphi},\; \rho > R \end{matrix}\right. \quad (18)$

We have used the fact that outside the cylindrical wire, we have a vacuum, hence

$\mathbf{B} = \mu_{0}\mathbf{H} \quad (19)$