Calculating the Magnetic Field in Free Space

[miew]

Homework Statement

If the electric field in free space is E=Eo(x^+y^)sin(2pi/lamda)(z+ct), with Eo=2 statvolts/cm the magnetic field, not including any static magnetic field, must be what?

2. Relevant equation

∇ x B= 1/c ∂E/∂t

The Attempt at a Solution

First I calculated ∂E/∂t= cE0/c (x^+y^)cos(2pi/lamba)(x+ct)

Since B is perpendicular to E, it just has a z^ direction.

∇xB=∂yBzx^-∂xBzj^.

Then I equal both equations and I got Bz=Boc(yx^-xj^) cos...

Is this right ?

 

[betel]

I assume your hats mean unit vectors? So your E field is propagating in the z direction and the Field vector points in $$\hat x + \hat y$$?

You made some mistake when you calculated $$\partial E/\partial t$$. The argument inside the sine should not change and you get some additional factors.
You first guess for B is wrong. Z is not the only direction perpendicular to E.

 

[miew]

Yes, they are unit vector. And the x inside the sin it is supposed to be a z, sorry :/

So aren't there two directions in which B can be perpendicular? x+z and y+z ?

 

[betel]

There are two directions but in your first post you gave only one, z. And they are not the ones you gave in your last post.
In addition you only calculated $$B_z=..\hat x$$ as proportional to a vector. But Bz should just be a number.

Unless you know from somewhere else that E and B are perpendicular, so far from the equations you don't see it. In this case you would have to start with a general B.

 

[miew]

Okay, I start with a general B.

So then,
∂yBz-∂zBy= Eocos(A) (where A is everything after cos)
∂zBx-∂xBz= Eocos(A)
∂xBy=∂yBx=0

Is that right ? if it is, how do I solve it ?

 

[betel]

Yes. The trick is to differentiate again and then combine the equations that you only have one B in the equation. Then you can integrate.

 

[miew]

Differentiate with respect to what ? :/

 

[betel]

x,y,z. You just have to find the right combination.

 

[miew]

Okay, I am going to try that !

Thanks :)