To investigate the stability of the motion described in the preceding question, evaluate the second derivatives of \(U\) at \(\rho=a, z=0\), and show that they may be written \(\begin{gathered} \frac{\partial^{2} U}{\partial \rho^{2}}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} \\ \frac{\partial^{2} U}{\partial \rho \partial z}=0, \quad \frac{\partial^{2} U}{\partial z^{2}}=-\frac{q^{2}}{m}\left[B_{z} \rho \frac{\partial B_{z}}{\partial \rho}\right]_{\rho=a, z=0} \end{gathered}\) (Hint: You will need to use the \(\varphi\) component of the equation \(\boldsymbol{\nabla} \wedge \boldsymbol{B}=\mathbf{0}\), and the fact that, since \(B_{\rho}=0\) for all \(\rho, \partial B_{\rho} / \partial \rho=0\) also.) Given that the dependence of \(B_{z}\) on \(\rho\) near the equilibrium orbit is described by \(B_{z} \propto(a / \rho)^{n}\), show that the orbit is stable if \(0

Step by step solution

01

Use the curl equation

To evaluate the second derivatives, we need to use the given hint, which involves the equation for the curl of the magnetic field \(\nabla \wedge \boldsymbol{B} = \mathbf{0}\). We are given that \(B_{\rho} = 0\) and \(\frac{\partial B_{\rho}}{\partial \rho} = 0\). In cylindrical coordinates, the curl equation can be written as: $$ \nabla \wedge \boldsymbol{B} = \left(\frac{1}{\rho} \frac{\partial B_z}{\partial \rho} - \frac{\partial B_{\phi}}{\partial z}\right) \boldsymbol{\rho} + \left(\frac{\partial B_{\rho}}{\partial z} - \frac{\partial B_z}{\partial z}\right) \boldsymbol{\phi} + \frac{1}{\rho}\left(\frac{\partial \left(\rho B_{\phi}\right)}{\partial \rho} - \frac{\partial B_{\rho}}{\partial \phi}\right)\boldsymbol{z} $$ The \(\phi\) component must be zero according to the hint: $$ \frac{\partial B_{\rho}}{\partial z} - \frac{\partial B_z}{\partial z} = 0 $$ Since \(B_{\rho} = 0\), $$ \frac{\partial B_z}{\partial z} = 0 $$

02

Evaluate the expressions

Now, we will evaluate the expressions: $$ \frac{\partial^{2} U}{\partial \rho^{2}}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} $$

03

Determine the dependence of \(B_z\) on \(\rho\)

We are given that the dependence of \(B_z\) on \(\rho\) near the equilibrium orbit is described by \(B_{z} \propto(a / \rho)^{n}\). Therefore, we can write: $$ B_z \sim \left(\frac{a}{\rho}\right)^n $$ Differentiating with respect to \(\rho\), we get: $$ \frac{\partial B_z}{\partial \rho} \sim -n \left(\frac{a}{\rho}\right)^{n+1} $$ Now, we substitute these expressions back into the second derivatives: $$ {\frac{\partial^{2} U}{\partial \rho^{2}}}_{\rho=a, z=0}=\frac{q^{2}}{m}\left[B_{z}\left(B_{z}+\rho \frac{\partial B_{z}}{\partial \rho}\right)\right]_{\rho=a, z=0} = \frac{q^{2}}{m} B_z^2 \left(1 - n\right) $$ and $$ \frac{\partial^{2} U}{\partial z^{2}}=-\frac{q^{2}}{m}\left[B_{z} \rho \frac{\partial B_{z}}{\partial \rho}\right]_{\rho=a, z=0} = -n \frac{q^2}{m} B_z^2 $$

04

Check the stability

For the motion to be stable, the second derivative of the potential energy with respect to \(\rho\) and \(z\) should be positive and negative, respectively. So we need to satisfy: $$ {\frac{\partial^{2} U}{\partial \rho^{2}}}_{\rho=a, z=0} > 0\quad \text{and} \quad \frac{\partial^{2} U}{\partial z^{2}}_{\rho=a, z=0} < 0 $$ From the expressions above, we have: $$ \frac{q^{2}}{m} B_z^2 \left(1 - n\right) > 0\quad \text{and} \quad -n \frac{q^2}{m} B_z^2 < 0 $$ Since \(\frac{q^2}{m} B_z^2 > 0\), we can deduce that \(1 > n > 0\). Therefore, for the circular orbit to be stable, the condition \(0<n<1\) must be satisfied.

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