Question

) Given, E ̅ = E ̅m sin(x) sin (t) a ̂y and H ̅ = E ̅m/µo cos(x) cos (t) a ̂z. Determine whether these fields satisfy Maxwell’s equations or not.

Answer 1

Let's substitute them to the Maxwell’s equations (with no charges and currents in vacuum):

$$\nabla\cdot\mathbf{D}= 0\\ \nabla\cdot\mathbf{B}=0\\ \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}\\ \nabla\times\mathbf{H}= \frac{\partial\mathbf{D}}{\partial t}$$

Considering the fact that $\mathbf{D} = \varepsilon_0\mathbf{E}$ and $\mathbf{B} = \mu_0\mathbf{H}$, obtain:

1.

$$\nabla\cdot\mathbf{D}= \varepsilon_0 \left ( \dfrac{\partial E_x}{\partial x} + \dfrac{\partial E_y}{\partial y}+ \dfrac{\partial E_z}{\partial z} \right)$$

As far as $\mathbf{E} = (0,E_m\sin(x)sin(t),0)$, y-coordinate does not depend on y, so the divergence is equal to 0: $\nabla\cdot\mathbf{D}=0$. The first equation is satisfied.

2. Similarlly, as far as $\mathbf{H} = (0,0,E_m/\mu_0 \cos(x), \cos(t))$, the z-coordinate does not depend on z, thus, the divergens will be equal to 0: $\nabla\cdot\mathbf{B} = \nabla\cdot\mathbf{H}=0$. The second equation is satisfied.

3. The curle is:

$$\nabla\times\mathbf{E}= \left(\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}\right) \mathbf e_x+ \left(\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}\right) \mathbf e_y+ \left(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}\right) \mathbf e_z$$

From all these derivatives only $\dfrac{\partial E_y}{\partial x}$ will not be equal to 0. Let's calculate it:

$$\dfrac{\partial E_y}{\partial x} =E_m\sin(t)\dfrac{\partial }{\partial x}\sin(x) = E_m\sin(t)\cos(x)$$

Thus:

$$\nabla\times\mathbf{E}= (0,0,E_m\sin(t)\cos(x) )$$

On the other hand:

$$-\dfrac{\partial\mathbf{B}}{\partial t} =-\mu_0\dfrac{\partial\mathbf{H}}{\partial t}= \left(0,0, - \dfrac{\partial}{\partial t} [E_m \cos(x), \cos(t)] \right)=\\ =(0,0, E_m\sin(t)\cos(x))$$

Thus, $\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}$ and the third equation is also satisfied.

4. The curle is:

$$\nabla\times\mathbf{H}= \left(\frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z}\right) \mathbf e_x+ \left(\frac{\partial H_x}{\partial z} - \frac{\partial H_z}{\partial x}\right) \mathbf e_y+ \left(\frac{\partial H_y}{\partial x} - \frac{\partial H_x}{\partial y}\right) \mathbf e_z$$

From all these derivatives only $-\dfrac{\partial H_z}{\partial x}$ will not be equal to 0. Let's calculate it:

$$-\dfrac{\partial H_z}{\partial x} =-E_m/\mu_0\cos(t)\dfrac{\partial }{\partial x}\cos(x) =E_m/\mu_0\cos(t)\sin(x)$$

Thus:

$$\nabla\times\mathbf{H}= \dfrac{1}{\mu_0} (0,E_m\cos(t)\sin(x),0)$$

On the other hand:

$$\dfrac{\partial\mathbf{D}}{\partial t} =\varepsilon_0\dfrac{\partial\mathbf{E}}{\partial t}= \varepsilon_0\left(0, \dfrac{\partial}{\partial t} [E_m \sin(x), \sin(t)],0 \right)=\\ =\varepsilon_0(0,E_m\cos(t)\sin(x),0)$$

Because of the coeffitients $\dfrac{1}{\mu_0}$ and $\varepsilon_0$ the right hand side is not equal to the left hand side:

$$\nabla\times\mathbf{H}\ne \frac{\partial\mathbf{D}}{\partial t}$$

Thus, the fourth equation is NOT satisfied.

Answer. These fields do not satisfy Maxwell’s equations.