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Chapter 8: Problem 2

A charged particle in an electromagnetic field experiences the Lorentz force $$ \mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}), $$ where \(q\) is the charge and \(\mathbf{v}\) the (vector) velocity of the particle. Show that an electromagnetic wave in free space acts on a charged particle primarily through its electric field, the magnetic interaction being smaller by at least the ratio \(|\mathbf{v}| / c\).

Short Answer

Expert verified
As an electromagnetic wave travels through free space, explain whether the magnetic field or electric field has a more significant effect on a charged particle and why. The electric field has a more significant effect on a charged particle when an electromagnetic wave travels through free space. This is due to the electric and magnetic forces acting on the charged particle. When comparing the magnitudes of these forces (electric force, \(\mathbf{F_E}\), and magnetic force, \(\mathbf{F_B}\)), the magnetic interaction is smaller than the electric interaction by a factor of the ratio \(|\mathbf{v}|/c\), where \(\mathbf{v}\) is the particle's velocity and \(c\) is the speed of light. This means that the charged particle experiences greater acceleration and force from the electric field compared to the magnetic field in free space.

Step by step solution

01

Calculate the Magnitudes of Electric and Magnetic Forces

First, we need to find the magnitudes of electric force \(\mathbf{F_E}\) and magnetic force \(\mathbf{F_B}\) experienced by the charged particle. From the Lorentz force equation, we know that the electric force and magnetic forces are given by: $$ \mathbf{F_E} = q\mathbf{E}, $$ and $$ \mathbf{F_B} = q(\mathbf{v} \times \mathbf{B}). $$
02

Calculate the Ratio of Electric and Magnetic Fields

We know that in an electromagnetic wave in free space, the magnitudes of electric field \(E\) and magnetic field \(B\) are related by the speed of light \(c\), which can be given by: $$ c = \frac{E}{B}. $$
03

Express \(\mathbf{F_B}\) in terms of Electric Field, Speed of Light and Charged Particle Velocity

Since we are interested in the comparison between \(\mathbf{F_E}\) and \(\mathbf{F_B}\), let's express the magnitude of the magnetic force in terms of the electric field. Using \(E = cB\), we can write: $$ \mathbf{F_B} = q(\mathbf{v} \times \frac{1}{c}\mathbf{E}), $$ which simplifies to $$ \mathbf{F_B} = \frac{q}{c}(\mathbf{v} \times \mathbf{E}). $$
04

Compare Magnitudes of \(\mathbf{F_E}\) and \(\mathbf{F_B}\)

Now let's compare the magnitudes of the electric and magnetic forces acting on the charged particle: $$ \frac{|\mathbf{F_B}|}{|\mathbf{F_E}|} = \frac{|\frac{q}{c}(\mathbf{v} \times \mathbf{E})|}{|q\mathbf{E}|}. $$ Notice that \(q\) cancels out on both sides, so we get: $$ \frac{|\mathbf{F_B}|}{|\mathbf{F_E}|} = \frac{|\mathbf{v} \times \mathbf{E}|}{c|\mathbf{E}|}. $$
05

Estimate the Smallest Value of the Ratio

Let's find the smallest value of this ratio. We know that the magnitude of the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is \(|\mathbf{a}||\mathbf{b}|\sin\theta\), where \(\theta\) is the angle between the vectors. In this case, \(\mathbf{a} = \mathbf{v}\) and \(\mathbf{b} = \mathbf{E}\), hence $$ \frac{|\mathbf{F_B}|}{|\mathbf{F_E}|} = \frac{|\mathbf{v}\sin\theta|}{c}. $$ The maximum value of \(\sin\theta\) is 1, so the smallest value of the above ratio is $$ \frac{|\mathbf{F_B}|}{|\mathbf{F_E}|} \ge \frac{|\mathbf{v}|}{c}. $$ Therefore, the magnetic interaction is smaller than the electric interaction by at least the ratio \(|\mathbf{v}|/c\).

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Most popular questions from this chapter

The text following (8.2.10) refers to low-frequency (or dc) laboratory measurements of \(\epsilon_{0}\) and \(\mu_{0}\). How could you determine these constants? What logical chain of definitions and calibrations would be needed?

Adapt the discussion at the end of Sec. 57 to show that the number of rectangular-waveguide modes whose cutoff frequencies are less than a given frequency \(\omega_{\max } x\) approximately, $$ N=\frac{\omega_{\max }^{2}}{2 \pi c^{2}} a b, $$ where \(N\) is assumed to be very large, and hence that the density of modes per unit frequency interval \(d N / d \omega\) increases linearly with frequency. Hint: Count both TE and TM modes.

Show, in general, that for TE modes the tangential-E boundary condition implies that the normal derivative of the scalar function \(\phi\) must vanish at the boundary, whereas for TM modes the normal-B boundary condition implies that the function \(\phi\) itself must vanish at the boundary.

Develop Poynting's theorem for the general material medium of relative permittivity \(\kappa_{\&}\) and permeability \(\kappa_{m}\) introduced in Prob. \(8.2 .2\); i.e., substitute the Maxwell curl equations \((8.2 .18)\) and \((8.2 .20)\) in the expansion of \(\nabla \cdot(\mathbf{E} \times \mathbf{H})\) to obtain $$ \oint_{S}(\mathbf{E} \times \mathbf{H}) \cdot d \mathbf{S}+\int_{V}\left(\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}+\mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}\right) d v+\int_{V} \mathbf{E} \cdot \mathbf{J} d v=0 $$ from which it follows that the Poynting vector is $$ s=\mathbf{E} \times \mathbf{H} $$ and the energy density is $$ \begin{aligned} \left(W_{1}\right)_{\text {leld }} &=\frac{1}{2} \mathbf{E} \cdot \mathbf{D}+\frac{1}{2} \mathbf{H} \cdot \mathbf{B} \\ &=\frac{1}{2} \kappa_{\ell} \epsilon_{0} E^{2}+\frac{1}{2 \kappa_{\mathrm{m}} \mu_{0}} B^{2} \end{aligned} $$ What restrictions on \(\kappa_{\&}\) and \(\kappa_{m}\) are necessary to obtain \((8.4 .24)\) ?

When matter is present, the phenomenon of polarization (electrical displacement of charge in a molecule or alignment of polar molecules) can produce unneutralized (bound) charge that properly contributes to \(\rho\) in \((8.2 .1)\). Similarly the magnetization of magnetic materials, as well as time-varying polarization, can produce efiective currents that contribute to \(J\) in \((8.24)\). These dependent source charges and currents, as opposed to the independent or "causal" free charges and currents, can be taken into account implicitly by introducing two new fields, the dectric displacement \(\mathbf{D}\) and the magnetic intensity \(\mathbf{H} .+\) For linear isotropic media, $$ \begin{aligned} &\mathbf{D}=\kappa_{\varepsilon} \epsilon_{0} \mathbf{E} \\ &\mathbf{H}=\frac{\mathbf{B}}{\kappa_{m} \mu_{0}} \end{aligned} $$ where \(\kappa_{0}\) is the relative permittivity (or dielectric constant) and \(\kappa_{m}\) is the rclative permeability of the medium. In this more general situation, Maxwell's equations are $$ \begin{aligned} &\nabla \cdot \mathbf{D}=\rho_{\text {ree }} \\ &\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t} \\ &\nabla \cdot \mathbf{B}=0 \\ &\nabla \times \mathbf{H}=\mathbf{J}_{\text {frea }}+\frac{\partial \mathbf{D}}{\partial l} \end{aligned} $$ Show that in a homogeneous material medium without free charges or currents, the fields obey the simple wave equation with a velocity of propagation $$ c^{\prime}=\frac{1}{\left(x_{q} \operatorname{tos}_{m} \mu_{0}\right)^{1 / 2}}=\frac{c}{\left(\alpha_{q} K_{m}\right)^{1 / 2}} $$ and that consequently the refractive index of the medium is given by $$ n=\left(x_{q} K_{m}\right)^{1 / 2} $$
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