Prove that the magnetic flux density, B, satisfies the wave equation.

Short Answer

Expert verified
Both the electric and magnetic flux densities, D and B, satisfy the wave equation by the application of vector calculus and Maxwell's equations, and therefore, the magnetic flux density B does as well.

Step by step solution

01

Start with Maxwell's equations

The useful Maxwell's equations here are \(\nabla × E = - \frac{\partial B}{\partial t}\)and \(\nabla × H = J + \frac{\partial D}{\partial t}\) . Here, D is electric flux density, E and H are electric and magnetic field strength respectively, J is current density and \( \frac{\partial}{\partial t} \) defines a partial derivative with respect to time.
02

Apply Vector Identity

Use the relation \(\nabla × (\nabla × A) = \nabla (\nabla \cdot A) - \nabla^2 A \) with A being E and B in respective Maxwell's equations.
03

Apply to Maxwell's equations

For \(\nabla × E\) (let's call this equation 1) and \(\nabla × H\) (equation 2), use the vector Identity. For equation 1, this gives \(\nabla × (\nabla × E) = -\nabla × (\frac{\partial B}{\partial t})\). And for equation 2, we get \(\nabla × (\nabla × H) = \nabla × J + \nabla × (\frac{\partial D}{\partial t})\)
04

Simplify equations

\(\nabla × (\nabla × E) = \nabla (\nabla \cdot E) - \nabla^2 E\). Since \(\nabla \cdot E = 0\) (another Maxwell's equation), drop off that term. Also, \(\nabla × (\nabla × H) = \nabla (\nabla \cdot H) - \nabla^2 H\). Since \(\nabla \cdot H = 0\), drop off that term here also. Substituting B for E and D for H in respective equations, we get \(-\nabla^2B = -\frac{\partial^2 B}{\partial t^2}\) and \(-\nabla^2D = \nabla ×J + \frac{\partial^2 D}{\partial t^2}\). Compared with general wave equation \(\nabla^2 \phi = \frac{\partial^2 \phi}{\partial t^2}\), both B and D are solutions of the wave equation.
05

Conclusion

It's proven that both magnetic and electric flux densities satisfy the wave equation through application of vector manipulation and Maxwell's equations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The line element in terms of the metric tensor, \(g_{\alpha \beta}\) is given by $$ d s^{2}=g_{\alpha \beta} d x^{\alpha} d x^{\beta}. $$ Show that the transformed metric for the transformation \(x^{\prime \alpha}=x^{\prime \alpha}\left(x^{\beta}\right)\) is given by $$ g_{\gamma \delta}^{\prime}=g_{\alpha \beta} \frac{\partial x^{\alpha}}{\partial x^{\prime} \gamma} \frac{\partial x^{\beta}}{\partial x^{\prime \delta}} $$

Let \(T^{\alpha}\) be a contravariant vector and \(S_{\alpha}\) be a covariant vector. a. Show that \(R_{\beta}=g_{\alpha \beta} T^{\alpha}\) is a covariant vector. b. Show that \(R^{\beta}=g^{\alpha \beta} S_{\alpha}\) is a contravariant vector.

For \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r=|\mathbf{r}|\), simplify the following. a. \(\nabla \times(\mathbf{k} \times \mathbf{r})\). b. \(\nabla \cdot\left(\frac{\mathrm{r}}{r}\right)\). c. \(\nabla \times\left(\frac{\mathbf{r}}{r}\right)\). d. \(\nabla \cdot\left(\frac{\mathrm{r}}{r^{3}}\right)\). e. \(\nabla \times\left(\frac{\mathrm{r}}{r^{3}}\right)\).

Prove the following vector identities: a. \((\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{d})-(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \cdot \mathbf{c})\) b. \((\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=(\mathbf{a} \cdot \mathbf{b} \times \mathbf{d}) \mathbf{c}-(\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}) \mathbf{d}\).

A particle moves on a straight line, \(\mathbf{r}=t \mathbf{u}\), from the center of a disk. If the disk is rotating with angular velocity \(\omega\), then \(\mathbf{u}\) rotates. Let \(\mathbf{u}=\) \((\cos \omega t) \mathbf{i}+(\sin \omega t) \mathbf{j}\) a. Determine the velocity, \(\mathbf{v}\). b. Determine the acceleration, a. c. Describe the resulting acceleration terms identifying the centripetal acceleration and Coriolis acceleration.

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free