/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Maxwell's equations relating the... [FREE SOLUTION] | 91Ó°ÊÓ

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Maxwell's equations relating the electric field \(\mathbf{E}\) and mag- netic field \(\mathbf{H}\) as they vary vith time in a region containing no charge and no current can be stated as follows: $$\begin{array}{ll}{\operatorname{div} \mathbf{E}=0} & {\operatorname{div} \mathbf{H}=0} \\ {\operatorname{curl} \mathbf{E}=-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}} & {\operatorname{curl} \mathbf{H}=\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}}\end{array}$$ where \(c\) is the speed of light. Use these equations to prove the following: $$\begin{array}{l}{\quad\left(\text { a) } \nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}\right.} \\ {\text { (b) } \nabla \times(\nabla \times \mathbf{H})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}} \\\ {\text { (c) } \nabla^{2} \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \quad[\text {Hint} \text { : Use Exercise } 27 .]} \\\ {\text { (d) } \nabla^{2} \mathbf{H}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}}\end{array}$$

Short Answer

Expert verified
Applying vector identities and substituting Maxwell's equations verify the given expressions for both \(\mathbf{E}\) and \(\mathbf{H}\).

Step by step solution

01

Use the Vector Identity

Recall the vector identity \((∇ × (∇ × \mathbf{F})) = ∇(∇⋅\mathbf{F})−∇²\mathbf{F}\) for any vector field \(\mathbf{F}\). Since \(\operatorname{div} \mathbf{E} = 0\) and \(\operatorname{div} \mathbf{H} = 0\), this simplifies to \((∇ × (∇ × \mathbf{E})) = −∇²\mathbf{E}\) and \((∇ × (∇ × \mathbf{H})) = −∇²\mathbf{H}\)."
02

Substitute for Curl of E

From Maxwell's equations, \(\operatorname{curl} \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}\). Substitute this into the expression for the curl of a curl: \((∇ × \operatorname{curl} \mathbf{E}) = ∇ × \left(-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}\right)\)."
03

Simplify the Expression Using Curl

Using the identity for the curl of a time derivative \(∇ × \left(-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}\right) = -\frac{1}{c} \frac{\partial}{\partial t}(∇ × \mathbf{H})\) and substituting \(\operatorname{curl} \mathbf{H} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}\), we obtain \(-\frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2}\)."
04

Prove Part (a) Conclusion

Hence, \(∇ × (∇ × \mathbf{E}) = -abla^2 \mathbf{E} = -\frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2}\). This confirms part (a)."
05

Substitute for Curl of H

Similarly, from Maxwell's equations, \(\operatorname{curl} \mathbf{H} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}\). Substitute this into the expression for the curl of a curl: \(∇ × (∇ × \mathbf{H}) = ∇ × \left(\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}\right)\)."
06

Simplify the Expression Using Curl

Using the similar steps as in Step 3, \(∇ × \left(\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}\right) = \frac{1}{c} \frac{\partial}{\partial t}(∇ × \mathbf{E})\), substitute \(\operatorname{curl} \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t}\) to reach \(-\frac{1}{c^2} \frac{\partial^2 \mathbf{H}}{\partial t^2}\)."
07

Prove Part (b) Conclusion

Hence, \(∇ × (∇ × \mathbf{H}) = -abla^2 \mathbf{H} = -\frac{1}{c^2} \frac{\partial^2 \mathbf{H}}{\partial t^2}\). This confirms part (b)."
08

Verify Wave Equations for E and H

Since \(-∇²\mathbf{E} = -\frac{1}{c^2} \frac{\partial^{2} \mathbf{E}}{\partial t^2}\), multiply through by -1, resulting in \(abla^2 \mathbf{E} = \frac{1}{c^2} \frac{\partial^{2} \mathbf{E}}{\partial t^2}\). The same applies to \(\mathbf{H}\), validating parts (c) and (d).

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics focusing on vector fields, crucial for understanding many physical phenomena. Here, vectors represent quantities with both magnitude and direction. Essential in electromagnetism, vector calculus uses operations like divergence, curl, and gradient to analyze vector fields like electric and magnetic fields.

- **Divergence** measures how much a vector field spreads out from a point. In Maxwell's equations for a charge-free space, both the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{H} \) have zero divergence, implying the absence of sources or sinks for these fields.
- **Curl**, on the other hand, measures the rotation or swirling of the field around a point. Maxwell's equations utilize curl to relate the changing electric and magnetic fields over time.

Intermediate concepts often use identities like \( abla \times ( abla \times \mathbf{F} ) = abla ( abla \cdot \mathbf{F} ) - abla^2 \mathbf{F} \). Applying this to divergence-free fields (where \( abla \cdot \mathbf{F} = 0 \)) results in simplified forms useful in deriving further electromagnetic properties. Vector calculus in this context serves to bridge mathematical theory with physical laws such as those governing electromagnetic waves.
Electromagnetic Waves
Electromagnetic waves are essential features of Maxwell's equations. These equations describe how electric and magnetic fields propagate through space. An electromagnetic wave consists of oscillating electric and magnetic fields that travel through a vacuum or a medium at the speed of light, \( c \).

- **Propagation**: The wave nature arises from the interdependence of electric and magnetic fields. When an electric field changes, it induces a magnetic field, and vice versa. This self-propagating nature is key in understanding how light and other electromagnetic radiation travel.
- **Characteristics** of electromagnetic waves include their speed, frequency, and wavelength. Maxwell's equations integrate these properties by relating the rate of change of one field with the curl of the other, allowing for the prediction and analysis of wave behavior in charge-free regions.

In classical electromagnetism, electromagnetic waves result from solutions to Maxwell's equations in free space. These solutions predict transverse waves (where fields are perpendicular to the direction of motion) with a fixed speed—light speed—emphasizing the intrinsic connection between electric and magnetic components.
Wave Equation
The wave equation is a fundamental aspect of physics describing how waves, such as electromagnetic waves, propagate through a medium. Specifically, for electric and magnetic fields, the wave equation stems from Maxwell's equations and takes the form \( abla^2 \mathbf{E} = \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} \) for the electric field \( \mathbf{E} \) and similarly for the magnetic field \( \mathbf{H} \).

- **Interpretation**: The wave equation indicates the relationship between spatial and temporal changes in the field. The \( abla^2 \) operator represents the spatial part, while the second-order time derivative denotes how the wave evolves over time.
- **Derivation**: Maxwell's equations, combined with vector calculus identities, lead to these wave equations. By recognizing that the divergence of \( \mathbf{E} \) and \( \mathbf{H} \) is zero, the curl-curl identity helps simplify and ultimately lead to the wave form.

The wave equation confirms that changes in electric or magnetic fields propagate as waves with a constant speed, \( c \). It models various physical scenarios—from light propagation to radio signal transmission—reinforcing its critical role in describing electromagnetic phenomena.

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Most popular questions from this chapter

Find the mass of a thin funnel in the shape of a cone \(z=\sqrt{x^{2}+y^{2}}, 1 \leqslant z \leqslant 4,\) if its density function is \(\rho(x, y, z)=10-z\).

\(21-27=\) Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If \(f\) is a scalar field and \(\mathbf{F},\) \(\mathbf{G}\) are vector fields, then \(f \mathbf{F}, \mathbf{F} \cdot \mathbf{G},\) and \(\mathbf{F} \times \mathbf{G}\) are defined by \((f \mathbf{F})(x, y, z)=f(x, y, z) \mathbf{F}(x, y, z)\) \((\mathbf{F} \cdot \mathbf{G})(x, y, z)=\mathbf{F}(x, y, z) \cdot \mathbf{G}(x, y, z)\) \((\mathbf{F} \times \mathbf{G})(x, y, z)=\mathbf{F}(x, y, z) \times \mathbf{G}(x, y, z)\) $$\operatorname{div}(\mathbf{F} \times \mathbf{G})=\mathbf{G} \cdot \operatorname{curl} \mathbf{F}-\mathbf{F} \cdot \operatorname{curl} \mathbf{G}$$

Evaluate the surface integral \(\iint_{s} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\begin{array}{l}{\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}, \quad S \text { is the boundary of the }} \\ {\text { solid half- cylinder } 0 \leqslant z \leqslant \sqrt{1-y^{2}}, 0 \leqslant x \leqslant 2}\end{array}$$

\(5-15\) " Use the Divergence Theorem to calculate the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S} ;\) that is, calculate the flux of \(\mathbf{F}\) across \(S .\) $$\mathbf{F}(x, y, z)=x^{2} \sin y \mathbf{i}+x \cos y \mathbf{j}-x z \sin y \mathbf{k}$$ \(S\) is the "fat sphere" \(x^{8}+y^{8}+z^{8}=8\)

(a) Use Stokes' Theorem to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x^{2} y \mathbf{i}+\frac{1}{3} x^{3} \mathbf{j}+x y \mathbf{k}\) and \(C\) is the curve of intersection of the hyperbolic paraboloid \(z=y^{2}-x^{2}\) and the cylinder \(x^{2}+y^{2}=1\) oriented counterclockwise as viewed from above. (b) Graph both the hyperbolic paraboloid and the cylinder with domains chosen so that you can see the curve \(C\) and the surface that you used in part (a). (c) Find parametric equations for \(C\) and use them to graph \(C .\)

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