- #1
OhNoYaDidn't
- 25
- 0
Hey guys, i just came across this on my classical physics course.
So, I'm given that: [tex]E(z, t) = {E_{0}}sin(wt)sin(kz)\widehat{x}[/tex], and I'm supposed to find an expression for the associated magnetic field B.
Usually, i just find the propagation direction, and do it's cross product with the direction of E, and then write it as [tex]\overrightarrow{B}(r,t)=\frac{1}{c}\widehat{k}\times \overrightarrow{E} [/tex]but in this case, it doesn't seem to be as straightforward.
I thought of using [tex]\triangledown \times \overrightarrow{E} = -\frac{\partial \overrightarrow{B}}{\partial t}[/tex], finding the curl of E, and then integrate with respect to time, this would give me:
[tex] B(z, t) = \frac{k}{w}{E_{0}}cos(kz)cos(wt)\widehat{y} [/tex]
Is this right?
Thank you.
So, I'm given that: [tex]E(z, t) = {E_{0}}sin(wt)sin(kz)\widehat{x}[/tex], and I'm supposed to find an expression for the associated magnetic field B.
Usually, i just find the propagation direction, and do it's cross product with the direction of E, and then write it as [tex]\overrightarrow{B}(r,t)=\frac{1}{c}\widehat{k}\times \overrightarrow{E} [/tex]but in this case, it doesn't seem to be as straightforward.
I thought of using [tex]\triangledown \times \overrightarrow{E} = -\frac{\partial \overrightarrow{B}}{\partial t}[/tex], finding the curl of E, and then integrate with respect to time, this would give me:
[tex] B(z, t) = \frac{k}{w}{E_{0}}cos(kz)cos(wt)\widehat{y} [/tex]
Is this right?
Thank you.