Chapter 5, Problem 5.1P

Starting with Maxwell’s equations for simple, charge free media, derive the Helmholtz equation for H.

To determine

The Helmholtz equation for magnetic field $H$ .

Answer to Problem 5.1P

The Helmholtz equation for magnetic field $H$ is ${\nabla }^{2}H=\text{μσ}\frac{\partial H}{\partial t}+\text{με}\frac{{\partial }^{2}H}{\partial t^2}$ .

Explanation of Solution

Given:

The Maxwell’s equations are simple and the media is free of any charge.

Concept used:

The equation of Ampere’s circuital law for simple and charge free media is shown below.

  $\nabla \times H=\text{σ}E+\text{ε}\frac{\partial E}{\partial t}$ ....... (1)

The equation of Faraday’s law is shown below.

  $\nabla \times E=-\text{μ}\frac{\partial H}{\partial t}$

Calculation:

The condition for charge free media is $\nabla \cdot H=0$ .

Take curl on both sides of equation (1) as shown below.

  $\begin{array}{c}\nabla \times \left(\nabla \times H\right)=\nabla \times \left(\text{σ}E+\text{ε}\frac{\partial E}{\partial t}\right)\\ =\nabla \times \left(\text{σ}E\right)+\nabla \times \left(\text{ε}\frac{\partial E}{\partial t}\right)\\ =\text{σ}\left(\nabla \times E\right)+\text{ε}\left(\nabla \times \frac{\partial E}{\partial t}\right)\end{array}$

Since, the right side of above equation is a position derivative that acts on time derivative, so the above equation can be written as,

  $\nabla \times \left(\nabla \times H\right)=\text{σ}\left(\nabla \times E\right)+\text{ε}\frac{\partial }{\partial t}\left(\nabla \times E\right)$

Substitute $-\text{μ}\frac{\partial H}{\partial t}$ for $\nabla \times E$ in above equation.

  $\begin{array}{c}\nabla \times \left(\nabla \times H\right)=\text{σ}\left(-\text{μ}\frac{\partial H}{\partial t}\right)+\text{ε}\frac{\partial }{\partial t}\left(-\text{μ}\frac{\partial H}{\partial t}\right)\\ =-\text{μσ}\frac{\partial H}{\partial t}-\text{με}\frac{\partial }{\partial t}\left(\frac{\partial H}{\partial t}\right)\\ =-\text{μσ}\frac{\partial H}{\partial t}-\text{με}\frac{{\partial }^{2}H}{\partial t^2}\end{array}$

The curl of curl of any vector is equal to the divergence and Laplacian, so the above equation can be written as,

  $\nabla \cdot H-{\nabla }^{2}H=-\text{μσ}\frac{\partial H}{\partial t}-\text{με}\frac{{\partial }^{2}H}{\partial t^2}$

Due to charge free medium the value of divergence of magnetic field $\nabla \cdot H$ is equal to $0$ .

Now, the above equation can be written as,

  $\begin{array}{c}-{\nabla }^{2}H=-\text{μσ}\frac{\partial H}{\partial t}-\text{με}\frac{{\partial }^{2}H}{\partial t^2}\\ {\nabla }^{2}H=\text{μσ}\frac{\partial H}{\partial t}+\text{με}\frac{{\partial }^{2}H}{\partial t^2}\end{array}$

The above equation is known as Helmholtz wave equation for magnetic field $H$ .

Therefore, the Helmholtz equation is ${\nabla }^{2}H=\text{μσ}\frac{\partial H}{\partial t}+\text{με}\frac{{\partial }^{2}H}{\partial t^2}$ .

Conclusion:

Thus, the Helmholtz equation for magnetic field $H$ is ${\nabla }^{2}H=\text{μσ}\frac{\partial H}{\partial t}+\text{με}\frac{{\partial }^{2}H}{\partial t^2}$ .