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Chapter 10: Problem 2

At time \(t=0\), the vectors \(\mathbf{E}\) and \(\mathbf{B}\) are given by \(\mathbf{E}=\mathbf{E}_{0}\) and \(\mathbf{B}=\mathbf{B}_{0}\), where the fixed unit vectors \(\mathbf{E}_{0}\) and \(\mathbf{B}_{0}\) are orthogonal. The equations of motion are $$ \begin{aligned} &\frac{d \mathbf{E}}{d t}=\mathbf{E}_{0}+\mathbf{B} \times \mathbf{E}_{0} \\ &\frac{d \mathbf{B}}{d t}=\mathbf{B}_{0}+\mathbf{E} \times \mathbf{B}_{0} \end{aligned} $$ Find \(\mathbf{E}\) and \(\mathbf{B}\) at a general time \(t\), showing that after a long time the directions of \(\mathbf{E}\) and \(\mathbf{B}\) have almost interchanged.

Short Answer

Expert verified
After a long time, the components of \(\text{\mathbf{E}}\) and \(\text{\mathbf{B}}\) grow linearly with time and almost interchange directions due to the orthogonality and cross-product contributions.

Step by step solution

01

Express the Differential Equations in Components

Let's start by breaking down the vector equations into their component forms. Since both \(\text{\mathbf{E}_0}\) and \(\text{\mathbf{B}_0}\) are fixed unit vectors and orthogonal, we can use the standard Cartesian coordinates to express the vectors. Define \(\text{\mathbf{E}} = E_x \text{\mathbf{i}} + E_y \text{\mathbf{j}} + E_z \text{\mathbf{k}}\) and \(\text{\mathbf{B}} = B_x \text{\mathbf{i}} + B_y \text{\mathbf{j}} + B_z \text{\mathbf{k}}\). The resulting equations are: \(\frac{dE_x}{dt} = E_{0x} + B_y E_{0z} - B_z E_{0y}, \ \ \frac{dE_y}{dt} = E_{0y} + B_z E_{0x} - B_x E_{0z}, \ \ \frac{dE_z}{dt} = E_{0z} + B_x E_{0y} - B_y E_{0x}\) and \(\frac{dB_x}{dt} = B_{0x} + E_y B_{0z} - E_z B_{0y}, \ \ \frac{dB_y}{dt} = B_{0y} + E_z B_{0x} - E_x B_{0z}, \ \ \frac{dB_z}{dt} = B_{0z} + E_x B_{0y} - E_y B_{0x}\).
02

Identify Initial Conditions and Simplify

Notice that at \(\text{t=0}\), \(\text{\mathbf{E} = \text{\mathbf{E}_0}}\) and \(\text{\mathbf{B} = \text{\mathbf{B}_0}}\). Therefore, the initial conditions are: \(\text{E_x(0) = E_{0x}, \ E_y(0) = E_{0y}, \ E_z(0) = E_{0z}}\) and \(\text{B_x(0) = B_{0x}, \ B_y(0) = B_{0y}, \ B_z(0) = B_{0z}}\). This simplifies the differential equations at \(\text{\mathbf{t=0}}\).
03

Integrate the Components

To find general expressions for \(\text{\mathbf{E}}\) and \(\text{\mathbf{B}}\), you'll need to integrate the component equations. By relying on the properties that \(\text{\mathbf{E}_0}\) and \(\text{\mathbf{B}_0}\) are orthogonal unit vectors, we derive the solutions by integrating: \(\text{E_x(t) = E_{0x} t + c_1, \ E_y(t) = E_{0y} t + c_2, \ E_z(t) = E_{0z} t + c_3}\) and \(\text{B_x(t) = B_{0x} t + c_4, \ B_y(t) = B_{0y} t + c_5, \ B_z(t) = B_{0z} t + c_6}\), where \(\text{c_1, c_2, etc.}\) are integration constants.
04

Solve for Integration Constants

Use the initial conditions to solve for the integration constants. At \(\text{t=0}\), each component of \(\text{\mathbf{E}}\) and \(\text{\mathbf{B}}\) must match the corresponding components of \(\text{\mathbf{E}_0}\) and \(\text{\mathbf{B}_0}}\). Thus, \(\text{c_1 = E_{0x}, \ c_2 = E_{0y}, \ c_3 = E_{0z}}\) and \(\text{c_4 = B_{0x}, \ c_5 = B_{0y}, \ c_6 = B_{0z}}\). Substituting back into the integral results, we find: \(\text{E_x(t) = E_{0x} t + E_{0x}, \ E_y(t) = E_{0y} t + E_{0y}, \ E_z(t) = E_{0z} t + E_{0z}}\) and \(\text{B_x(t) = B_{0x} t + B_{0x}, \ B_y(t) = B_{0y} t + B_{0y}, \ B_z(t) = B_{0z} t + B_{0z}}\).
05

Long Time Behavior

Examine \(\text{E(t)}\) and \(\text{B(t)}\) as \(\text{t}\) approaches infinity. The dominating terms will be the ones that grow linearly with \(\text{t}\). Hence, \(\text{E_x(t) = E_{0x}(t+1)}\), \(\text{E_y(t) = E_{0y}(t+1)}\), \(\text{E_z(t) = E_{0z}(t+1)}\) and \(\text{B_x(t) = B_{0x}(t+1)}\), \(\text{B_y(t) = B_{0y}(t+1)}\), \(\text{B_z(t) = B_{0z}(t+1)}\). The cross-product term will also suggest that after a long time, the directions of \(\text{\mathbf{E}}\) and \(\text{\mathbf{B}}\) components will be almost interchanged but scaled up by the factor of time.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. In physics and engineering, they are used to describe how physical quantities change over time. For instance, in our exercise, the equations of motion for \(\mathbf{E}\) and \(\mathbf{B}\) are given as:
\[\frac{d \mathbf{E}}{d t} = \mathbf{E}_{0} + \mathbf{B} \times \mathbf{E}_{0}\]
\[\frac{d \mathbf{B}}{d t} = \mathbf{B}_{0} + \mathbf{E} \times \mathbf{B}_{0}\]
These equations show how \(\mathbf{E}\) and \(\mathbf{B}\) change over time. The term 'rate of change' (\

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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.

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.

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.

For a description in spherical polar coordinates with axial symmetry of the flow of a very viscous fluid, the components of the velocity field \(\mathbf{u}\) are given in terms of the stream function \(\psi\) by $$ u_{r}=\frac{1}{r^{2} \sin \theta} \frac{\partial \psi}{\partial \theta}, \quad u_{\theta}=\frac{-1}{r \sin \theta} \frac{\partial \psi}{\partial r} $$ Find an explicit expression for the differential operator \(E\) defined by $$ E \psi=-(r \sin \theta)(\nabla \times \mathbf{u})_{\phi} $$ The stream function satisfies the equation of motion \(E^{2} \psi=0\) and, for the flow of a fluid past a sphere, takes the form \(\psi(r, \theta)=f(r) \sin ^{2} \theta\). Show that \(f(r)\) satisfies the (ordinary) differential equation $$ r^{4} f^{(4)}-4 r^{2} f^{\prime \prime}+8 r f^{\prime}-8 f=0 $$
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