Answered: in the space E(x,y;t)=…

Related questions

Question

in the space E(x,y;t)= az0.1cos(10πx)sin(6π109t-βy) (V/m).

find H(x,y;t) and β

electromagnetic theory

Expert Solution

Step 1

Given,

Electric field in space

Step 2

Accroding to maxwell equation

$μ_{0} \frac{\partial H}{\partial t} = - (\nabla \times E) μ_{0} \frac{\partial H}{\partial t} = - |\begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & 0.1 \cos (10 π x) \sin (6 π * 10^{9} t - β y) \end{matrix}| μ_{0} \frac{\partial H}{\partial t} = i (\frac{\partial (0.1 \cos (10 π x) \sin (6 π * 10^{9} t - β y))}{\partial y}) - j (\frac{\partial (0.1 \cos (10 π x) \sin (6 π * 10^{9} t - β y))}{\partial x}) μ_{0} \frac{\partial H}{\partial t} = (- 0.1 β \cos (10 π x) \cos (6 π * 10^{9} t - β y)) \hat{a_{x}} + (π \sin (10 π x) \sin (6 π * 10^{9} t - β y)) \hat{a_{y}} \frac{\partial H}{\partial t} = [\frac{(- 0.1 β \cos (10 π x) \cos (6 π * 10^{9} t - β y)) \hat{a_{x}} + (π \sin (10 π x) \sin (6 π * 10^{9} t - β y)) \hat{a_{y}}}{μ_{0}}] H = \int [\frac{(- 0.1 β \cos (10 π x) \cos (6 π * 10^{9} t - β y)) \hat{a_{x}} + (π \sin (10 π x) \sin (6 π * 10^{9} t - β y)) \hat{a_{y}}}{μ_{0}}] d t H = \frac{(- 0.1 β \cos (10 π x) \sin (6 π * 10^{9} t - β y)) \hat{a_{x}}}{μ_{0} * 6 π * 10^{9}} - \frac{(π \sin (10 π x) \cos (6 π * 10^{9} t - β y)) \hat{a_{y}}}{μ_{0} * 6 π * 10^{9}}$

Step by step

Solved in 3 steps