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Physics Bootcamp

Samuel J. Ling

Section 42.5 Maxwell’s Equations Bootcamp

Exercises Exercises

Maxwell’s Displacement Current

1. Displacement Current in a Charging Capacitor.
Follow the link: Example 42.3.
2. Computing Displacement Current Density and Current Between Plates of a Parallel Capacitor.
Follow the link: Example 42.4.
3. Displacement Current from a Quadratically Varying Voltage Across Capacitor Plates.
Follow the link: Exercise 42.1.1.
4. Displacement Current Density in a Region of Space Where Electric Field is Varying with Time.
Follow the link: Exercise 42.1.2.

Electromagnetic Waves

10. Compute Various Properties of an Electromagnetic Plane Wave Given its Electric Field.
Follow the link: Exercise 42.4.9.2.
11. Compute Various Properties of an Electromagnetic Plane Wave Given its Magnetic Field.
Follow the link: Exercise 42.4.9.3.
12. Electromagnetic Wave Function from Given Amplitude of Electric Wave and Frequency.
Follow the link: Exercise 42.4.9.4.
13. Electromagnetic Wave Function from Given Amplitude of Magnetic Wave and Frequency.
Follow the link: Exercise 42.4.9.5.
14. Electromagnetic Wave Function from Magnetic Wave, Frequency, and Phase.
Follow the link: Exercise 42.4.9.6.

Miscellaneous

17.
The magnetic field of a steady current volume density of magnitude \(J_0\) pointed in positive \(z\)-axis in a very long wire of radius \(R\) is given in polar coordinates as follows.
\begin{equation*} \vec B = \left\{ \begin{array}{ll} \frac{ {\displaystyle \mu_0 J_0} }{ {\displaystyle 2} }\; r\; \hat u_{\phi} \amp \ \ \ \ r\le R\\ \mbox{ } \amp \mbox{ }\\ \frac{ {\displaystyle \mu_0 J_0} }{ {\displaystyle 2} }\; \frac{{\displaystyle R^2 }}{ {\displaystyle r }}\; \hat u_{\phi} \amp \ \ \ \ r \gt R \end{array} \right. \end{equation*}
Here \(r\) is the distance from the wire and \(\hat u_{\phi}\) is a unit vector tangent to the circle centered at the wire of radius \(r\) passing through the field point. (a) Find the curl of the given magnetic field. (b) Prove that for a steady current \(\vec{\nabla}\times\vec B = \mu_0 \vec J\text{.}\) [Hint: You can write the given magnetic field in Cartesian coordinates and then calculate the curl.]
Hint.
The curl will depend on the point \(r\text{.}\) For points for which \(r \gt R\) the curl will be zero. For points \(r \lt R\) the curl will be equal to \(\mu_0\vec J\text{.}\) (b) You may start from the general formula for the magnetic field of a steady current given by the Biot-Savart integral and apply the curl to that magnetic field. This part requires advanced mathematics for some students and is left a challenge for more enterprising students.
18.
According to the Maxwell’s equations, a dynamic electric field that is uniform in \(y\) and \(z\) coordinates but allowed to vary along the \(x\)-axis obeys the following equations in free space, i.e. at points in space where there are no charges or currents.
\begin{equation*} E_x = 0,\quad \frac{\partial^2 E_y}{\partial t^2} = c^2 \frac{\partial^2 E_y}{\partial x^2},\quad \frac{\partial^2 E_z}{\partial t^2} = c^2 \frac{\partial^2 E_z}{\partial x^2}. \end{equation*}
where \(c^2 = 1/\mu_0\epsilon_0\text{.}\) Consider the following wave solution of these equations.
\begin{equation*} \vec E = \vec E_0 \cos(kx - \omega t), \end{equation*}
where \(\vec E_0\) is a constant vector in space whose \(x\)-component is zero. (a) Show that the components of the assumed solution satisfy the equations above. (b) Find the direction of motion of the two components of the electric field. (c) Write the associated magnetic field.
Hint.
(a) Write out the components of the given wave solution and plug into the wave equations for those components. (b) Examine the argument of the cosine. (c) Use \(\vec E \times \vec B\) to be in the direction of travel of the wave to find the direction of the magnetic field vector.
19.
The following represents an electromagnetic wave traveling in the direction of the positive \(x\)-axis.
\begin{equation*} E_x = 0;\ E_y = E_0 \cos(k x - \omega t);\ E_z = 0. \end{equation*}
\begin{equation*} B_x = 0;\ B_y = 0; B_z= B_0 \cos(k x - \omega t). \end{equation*}
The wave is passing through a wide tube of circular cross-section of radius \(R\) whose axis is along the \(x\)-axis. Find the expression for the displacement current through the tube
Solution.
The displacement current density at any point in space is given by
\begin{equation*} \vec J_d = \epsilon_0\:\dfrac{\partial \vec E}{\partial t}. \end{equation*}
Therefore, we find the following for the displacment current density at a point inside the tube.
\begin{equation*} \vec J_d = \hat u_y\:\epsilon_0\omega\: \: E_0\:\sin(k x - \omega t). \end{equation*}
Now, integrating this over a cross-section of the tube gives the following for the displacement current through th etub.
\begin{equation*} I_d = \pi\epsilon_0\omega\: R^2\: E_0\:\sin(k x - \omega t) \end{equation*}
20.
For scalar field \(\phi\) and vector field \(\vec A\text{,}\) prove the following identities for the del vector operator.
\begin{align*} \text{(a)}\quad \amp \vec{\nabla}\times(\vec{\nabla}\phi) = 0.\\ \text{(b)}\quad \amp \vec{\nabla}\cdot(\vec{\nabla}\times \vec A) = 0.\\ \text{(c)}\quad \amp \vec{\nabla}\times(\vec{\nabla}\times \vec A) = \vec{\nabla}(\vec{\nabla}\cdot\vec A) - (\vec{\nabla}\cdot \vec{\nabla})\vec A. \end{align*}
Hint.
Prove by actually calculating the given expressions using Cartesian components. For (a) and (c) you may show the relation to be true for the \(x\)-component of the equation since the same will be true for the other components as well.
21.
The electric field of an electromagnetic wave is given by the superposition of two waves.
\begin{equation*} \vec E(x,y,z,t) = \hat u_y E_0 \cos(kx-\omega t) + \hat u_y E_0 \cos(kx+\omega t). \end{equation*}
You can see that the first wave is moving towards the positive \(x\)-axis and the second moving towards the negative \(x\)-axis. (a) What is the associated magnetic field wave? (b) Write the Poynting vector and calculate the average energy per unit area per unit time transported by this wave.
Solution 1. a
(a) We work out the associated magnetic field wave for each of the two component waves in the given wave function. Since \(\hat u_y E_0 \cos(kx-\omega t)\) moves towards the positive \(x\)-axis, the associated magnetic field for this wil be \(\hat u_z \:\dfrac{E_0}{v}\:\cos(kx-\omega t)\text{,}\) where \(v\) is the speed of the wave in the given medium. If the medium is the vacuum, then \(v=c\text{.}\) The other part \(\hat u_y E_0 \cos(kx+\omega t)\) moves towards the negative \(x\)-axis, therefore the associated magnetic field wave will be \(-\hat u_z \:\dfrac{E_0}{v}\ \cos(kx+\omega t)\text{.}\) Therefore, the net magnetic field wave will be
\begin{equation*} \vec B(x,y,z,t) = \hat u_z \:\dfrac{E_0}{v}\ \cos(kx-\omega t) -\hat u_z \:\dfrac{E_0}{v}\ \cos(kx+\omega t). \end{equation*}
Solution 2. b
The Poynting vector is
\begin{equation*} \vec S = \dfrac{1}{\mu_0}\:\vec E\times \vec B = \hat u_x\:\:\dfrac{1}{\mu_0}\:\dfrac{E_0^2}{v}\:\left[\cos^2(kx-\omega t) -\cos^2(kx+\omega t)\right]. \end{equation*}
The quantity in \([\cdots]\) can be written as product of two trig functions and is left as an exercise.
22.
A microscopic spherical dust particle of radius 2 \(\mu\)m and mass 10 \(\mu\)g is moving in outer space at a constant speed of 30 cm/sec. A wave of light strikes it from the opposite direction of its motion, and gets absorbed. Assuming the particle decelerates uniformly to zero speed in one second, what is the average electric field amplitude in the light?
Answer.
\(7.35 \times 10^6\, \textrm{N/C}\text{.}\)
23.
A Styrofoam spherical ball of radius 2 mm and mass 20 \(\mu\)g is to be suspended by the radiation pressure in a vacuum tube in a lab. How much intensity will be required if the light is completely absorbed the ball?
Answer.
\(4.7 \times 10^6 \ \textrm{W/m}^2\text{.}\)
Solution.
Since the ball absorbs the radiation completely, the radiation pressure on the ball will be
\begin{equation*} P = \dfrac{I}{c}, \end{equation*}
where \(I\) is the intensity of the EM-wave and \(c\) its speed. In the present situation there will be an upward force by the radiation pressure that will balance the weight of the ball. We obtain the magnitude of the radiation force by multiplying the radiation pressure by the area of cross-section of the ball that presents itself in the path of the radiation, which is \(\pi R^2\text{,}\) where \(R\) is the radius of the ball. Therefore,
\begin{equation*} \dfrac{I}{c}\:\times \pi R^2 = m\:g\ \ \Longrightarrow\ \ I = \dfrac{ m\:g\:c}{\pi R^2 }. \end{equation*}
Now, putting in the numerical values we obtain
\begin{equation*} I = \dfrac{ 20\times 10^{-9}\:\textrm{kg}\times 9.8\:\textrm{m/s}^2:\times 3\times 10^8\:\textrm{m/s}}{\pi (0.002\:\textrm{m})^2 }= 4.7\times 10^6\:\textrm{W/m}^2. \end{equation*}
24.
A conservative force is given by the negative of the gradient of a potential energy function \(U\text{.}\) Based on the following given potential energy when the masses are separated by a distance \(r\) deduce the expression of the gravitational force on a mass \(m\) by another mass \(M\text{.}\)
\begin{equation*} U = -\frac{G_N M m }{r}\ \ \textrm{(Reference at )} \end{equation*}
Solution.
The conservative force \(\vec F\) and the potential energy \(U\) due to this force are related as follows.
\begin{equation*} \vec F = - \vec \nabla U. \end{equation*}
Therefore, we need to find the negative gradient of the given potential energy expression.
\begin{equation*} \vec F =G_N M m \vec \nabla \left(\frac{1}{r}\right) = -G_N M m \left(\frac{1}{r^2}\right)\: \hat u_r, \end{equation*}
where \(\hat u_r\) is a unit vector in the direction from \(M\) at the origin to \(m\text{,}\) which is located at a distance \(r\) from \(M\text{.}\)
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