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Chapter 10: Problem 17

The (Maxwell) relationship between a time-independent magnetic field \(\mathbf{B}\) and the current density \(\mathbf{J}\) (measured in S.I. units in \(\mathrm{A} \mathrm{m}^{-2}\) ) producing it, $$ \nabla \times \mathbf{B}=\mu_{0} \mathbf{J} $$ can be applied to a long cylinder of conducting ionised gas which, in cylindrical polar coordinates, occupies the region \(\rhoa\) and \(B=B(\rho)\) for \(\rhoa\). Like \(\mathbf{B}\), the vector potential is continuous at \(\rho=a\) (c) The gas pressure \(p(\rho)\) satisfies the hydrostatic equation \(\nabla p=\mathbf{J} \times \mathbf{B}\) and vanishes at the outer wall of the cylinder. Find a general expression for \(p\)

Short Answer

Expert verified
Current density \(C\) determined, Magnetic field \(B beliefved in cylinder; \mathbf{A} matches; Pressure \( \mathbf{J} \times \mathbf{B}.

Step by step solution

01

Understand Given Conditions

We have a current density \(\mathbf{J} = (0, C, 0)\) and a magnetic field \(\mathbf{B} = (0, 0, B)\), with \( B = B_0 \) for \( \rho > a \) and \( B = B(\rho) \) for \( \rho < a \). We need to show these are consistent with the Maxwell equation \(\abla \times \mathbf{B} = \mu_0 \mathbf{J} \).
02

Calculate \( abla \times \mathbf{B} \)

In cylindrical coordinates, the curl is given by \[abla \times \mathbf{B} = \left( \frac{1}{\rho} \frac{\partial B_\theta}{\partial z} - \frac{\partial B_z}{\partial \theta}, \frac{\partial B_z}{\partial \rho} - \frac{\partial B_\rho}{\partial z}, \frac{1}{\rho} \left( \frac{\partial (\rho B_\theta)}{\partial \rho} - \frac{\partial B_\rho}{\partial \theta} \right) \right]\].
03

Evaluate \( abla \times \mathbf{B} \) for the Given \( \mathbf{B} \)

Since \( \mathbf{B} = (0, 0, B(\rho)) \), most terms in the curl equation will vanish, leaving only \[\abla \times \mathbf{B} = \left( 0, -\frac{\partial B(\rho)}{\partial \rho}, 0 \right)\].
04

Apply Maxwell's Equation

From \( \abla \times \mathbf{B} = \mu_0 \mathbf{J} \), equate the result:\[(0, -\frac{\partial B(\rho)}{\partial \rho}, 0) = \mu_0(0, C, 0)\].
05

Solve for \( C \)

Thus, \(-\frac{\partial B(\rho)}{\partial \rho} = \mu_0 C\rightarrow\ C = -\mu_0 \frac{\partial B(\rho)}{\partial \rho}.\)
06

Continuity at \( \rho = a \)

Since \( \mathbf{B} \) is continuous at \( \rho = a \), \( B(a) = B_0 \), integrating \( B \) inside gives\[B(\rho) = B_0 + \mu_0 C (\rho - a)\].
07

Finding \( \mathbf{A} \)

To show a suitable \( \mathbf{A} \) with only one non-vanishing component, use \[\mathbf{B} = \abla \times \mathbf{A}.\]
08

Express \( \mathbf{A} \( for \( \rho < a \)

Choose \[\mathbf{A} = (0, 0, A_\phi(\rho)\]and \( \abla \times \mathbf{A} = (0, 0, \frac{1}{\rho} \frac{d(\rho A_\phi)}{d \rho}) \) to match with \( B(\rho) \).
09

Determine \( A_\phi(\rho) \)

Inside \( \rho < a \), \[B(\rho) = \frac{1}{\rho} \frac{d(\rho A_\phi)}{d \rho},A_\phi = -\frac{\mu_0 C}{2}\rho + \alpha/ \rho\].
10

Match Continuity at \( \rho = a \)

Ensure \[A_\phi = B_0\,, Solve \alpha term accordingly.\]
11

Express Gas Pressure

From \( \abla p = \mathbf{J} \times \mathbf{B} \), integrate \[p = \frac{C ((B_0))^2-{\rho^2}}{2}.\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current Density
Current density, denoted as \(\mathbf{J}\), refers to the amount of electric current flowing through a unit area of a material. In this exercise, the current density is given as \(\mathbf{J} = (0, C, 0)\), which means it is only present in the \(\theta\) direction (azimuthal direction) and is uniform. Current density is vital because it directly relates to the generation of magnetic fields according to Maxwell's equations.
The relevant Maxwell's equation used here is \(abla \times \mathbf{B}=\mu_{0} \mathbf{J}\), which states that the curl of the magnetic field \(\mathbf{B}\) is proportional to the current density \(\mathbf{J}\). Using this equation helps determine the characteristics of the magnetic field generated by a given current density.
By applying this to cylindrical coordinates, we can analyze how a uniform current density in a cylindrical conductor affects the magnetic field within and outside the conductor. Understanding this relationship is key to solving the given exercise problem.
Magnetic Field Continuity
Magnetic field continuity is a crucial principle stating that the magnetic field \(\mathbf{B}\) must be continuous across any boundary, i.e., it cannot suddenly change value. In the context of this exercise, it means that at the boundary \(\rho = a\) (surface of the cylinder), the magnetic field inside the cylinder must match the magnetic field outside.
Given the magnetic field \(\mathbf{B} = (0, 0, B)\), with \(B = B_0\) for \(\rho > a\) (outside) and \(B = B(\rho)\) for \(\rho < a\) (inside), continuity requires that \(B(a) = B_0\). This is essential when solving for the expression of \(B(\rho)\) within the cylinder because it defines the boundary condition used in integration and determining the constants.
  • Inside the cylinder (\rho < a), the current density generates a magnetic field that changes with \(\rho\).
  • Outside the cylinder (\rho > a), the magnetic field remains constant and equals \(B_0\).
Integrating these surrounding magnetic field conditions ensures the continuity principle is upheld.
Vector Potential
The vector potential \(\mathbf{A}\) is a vector field whose curl gives the magnetic field, expressed as \(\mathbf{B}=abla \times \mathbf{A}\). The vector potential simplifies calculations in electromagnetism, especially when solving Maxwell's equations.
In cylindrical coordinates for this exercise, \(\mathbf{A}\) is chosen to have only one non-vanishing component \(A_{\phi}\)(\rho), which drastically simplifies the math. To find \(A_{\phi}\), we consider:
  • For \(\rho < a\) (inside): The expression derived is \(A_{\phi}= - \frac{\mu_0 C}{2}\rho + \frac{\alpha}{\rho}\).
  • For \(\rho > a\) (outside): The constant \(B_0\) makes \(A_{\phi}= \frac{B_0 \rho}{2}\).
These forms ensure that \(\mathbf{B}\) derived from \(\mathbf{A}\) matches the given conditions and satisfies the boundary continuity at \(\rho=a\).
The continuity of \(\mathbf{A}\) at \(\rho = a\) translates to ensuring the forms inside and outside the boundary region match in value, which solves for the constant \(\alpha\).
Hydrostatic Equation
The hydrostatic equation in the context of electromagnetism describes how the pressure gradient in a conducting fluid balances the forces from the current density and magnetic field. Given by \(abla p=\mathbf{J} \times \mathbf{B}\), it relates the fluid pressure \(p(\rho)\) to the cross-product of current density and magnetic field.
From this exercise:
  • The gas pressure must balance the force generated by the current density and magnetic field, yielding \(abla p = \mathbf{J} \times \mathbf{B}\).
  • Given that \(\mathbf{J} = (0, C, 0)\) and \(\mathbf{B} = (0, 0, B(\rho))\), solving this cross-product leads to \(abla p= C \cdot B(\rho)\).
Integrate this equation starting from the boundary where the pressure is zero to find the expression for \(p(\rho)\) inside the cylinder. Given the boundary condition \(p(a) = 0\), the pressure inside can be written as \(p(\rho) = \frac{CB_{0}^2}{2}(a^2- \rho^2)\), representing how the pressure varies radially due to magnetic and current interactions.

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Most popular questions from this chapter

For the twisted space curve \(y^{3}+27 a x z-81 a^{2} y=0\), given parametrically by $$ x=a u\left(3-u^{2}\right), \quad y=3 a u^{2}, \quad z=a u\left(3+u^{2}\right) $$ show the following: (a) that \(d s / d u=3 \sqrt{2} a\left(1+u^{2}\right)\), where \(s\) is the distance along the curve measured from the origin; (b) that the length of the curve from the origin to the Cartesian point \((2 a, 3 a, 4 a)\) is \(4 \sqrt{2 a}\) (c) that the radius of curvature at the point with parameter \(u\) is \(3 a\left(1+u^{2}\right)^{2}\); (d) that the torsion \(\tau\) and curvature \(\kappa\) at a general point are equal; (e) that any of the Frenet-Serret formulae that you have not already used directly are satisfied.

Prove that for a space curve \(\mathbf{r}=\mathbf{r}(s)\), where \(s\) is the arc length measured along the curve from a fixed point, the triple scalar product $$ \left(\frac{d \mathbf{r}}{d s} \times \frac{d^{2} \mathbf{r}}{d s^{2}}\right) \cdot \frac{d^{3} \mathbf{r}}{d s^{3}} $$ at any point on the curve has the value \(\kappa^{2} \tau\), where \(\kappa\) is the curvature and \(\tau\) the torsion at that point.

If two systems of coordinates with a common origin \(O\) are rotating with respect to each other, the measured accelerations differ in the two systems. Denoting by \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) position vectors in frames \(O X Y Z\) and \(O X^{\prime} Y^{\prime} Z^{\prime}\) respectively, the connection between the two is $$ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \omega \times \dot{\mathbf{r}}+\omega \times(\omega \times \mathbf{r}) $$ where \(\omega\) is the angular velocity vector of the rotation of \(O X Y Z\) with respect to \(O X^{\prime} Y^{\prime} Z^{\prime}\) (taken as fixed). The third term on the RHS is known as the Coriolis acceleration, whilst the final term gives rise to a centrifugal force. Consider the application of this result to the firing of a shell of mass \(m\) from a stationary ship on the steadily rotating earth, working to the first order in \(\omega\left(=7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\right)\). If the shell is fired with velocity \(\mathbf{v}\) at time \(t=0\) and only reaches a height that is small compared to the radius of the earth, show that its acceleration, as recorded on the ship, is given approximately by $$ \ddot{\mathbf{r}}=\mathbf{g}-2 \omega \times(\mathbf{v}+\mathbf{g} t) $$ where \(m \mathbf{g}\) is the weight of the shell measured on the ship's deck. The shell is fired at another stationary ship (a distance \(\mathbf{s}\) away) and \(\mathbf{v}\) is such that the shell would have hit its target had there been no Coriolis effect. (a) Show that without the Coriolis effect the time of flight of the shell would have been \(\tau=-2 \mathbf{g} \cdot \mathbf{v} / g^{2}\) (b) Show further that when the shell actually hits the sea it is off target by approximately $$ \frac{2 \tau}{g^{2}}[(\mathbf{g} \times \omega) \cdot \mathbf{v}](\mathbf{g} \tau+\mathbf{v})-(\omega \times \mathbf{v}) \tau^{2}-\frac{1}{3}(\omega \times \mathbf{g}) \tau^{3} $$ (c) Estimate the order of magnitude \(\Delta\) of this miss for a shell for which \(v=300\) \(\mathrm{m} \mathrm{s}^{-1}\), firing close to its maximum range ( \(\mathbf{v}\) makes an angle of \(\pi / 4\) with the vertical) in a northerly direction, whilst the ship is stationed at latitude \(45^{\circ}\) North.

The shape of the curving slip road joining two motorways that cross at right angles and are at vertical heights \(z=0\) and \(z=h\) can be approximated by the space curve $$ \mathbf{r}=\frac{\sqrt{2} h}{\pi} \ln \cos \left(\frac{z \pi}{2 h}\right) \mathbf{i}+\frac{\sqrt{2} h}{\pi} \ln \sin \left(\frac{z \pi}{2 h}\right) \mathbf{j}+z \mathbf{k} $$ Show that the radius of curvature \(\rho\) of the curve is \((2 h / \pi) \operatorname{cosec}(z \pi / h)\) at height \(z\) and that the torsion \(\tau=-1 / \rho\). (To shorten the algebra, set \(z=2 h \theta / \pi\) and use \(\theta\) as the parameter.)

Evaluate the Laplacian of the function $$ \psi(x, y, z)=\frac{z x^{2}}{x^{2}+y^{2}+z^{2}} $$ (a) directly in Cartesian coordinates, and (b) after changing to a spherical polar coordinate system. Verify that, as they must, the two methods give the same result.
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