Obviously I'm seeing this a bit late to help the person who originally posted, but in case anybody else has issues...
The post before this is correct - the easiest way to calculate an EM wave's magnetic field from its electric field, or vice versa, is to remember that energy is shared equally between the fields.
[tex]E=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}B[/tex]
[tex]\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}[/tex] is the speed of light,
c, which is equal to [tex]3.00\times10^{8} m/s[/tex].
[tex]E=cB[/tex]
Answering the first two questions should be simple here. Since it doesn't appear that you are being asked for a direction/phase sign, you shouldn't need to use the right-hand rule.
For the last question, consider the amplitude of E. You are told that the maximum E-field is equal to 1000 V/m. This corresponds to a maximum of a sinusoidal wave with an amplitude of 1000 units. You are asked to find the shortest distance
on the wave between this point and a point where the E-field is 500 V/m, or half the amplitude. Since E-fields and B-fields are not measured in meters, you are really being asked to find this distance
on the axis of propagation. If we take the simple unit circle of circumference [tex]2\pi[/tex], and find the distance on the circle between the point where the y-value is 1 (the maximum) and a point where the y-value is 1/2 (half of the maximum), we calculate the length of that section of the circle is
[tex]\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}[/tex]
which is 1/6 of the total circumference. Since one turn of the circle represents one complete wavelength, and we know that the frequency of the wave is [tex]1.0\times10^{6}s^{-1}[/tex], we can calculate
[tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}ms^{-1}}{1.0\times10^{6}s^{-1}}=300m[/tex]
1/6 of the total wavelength is 50m.
Sorry for the wordiness - a picture would have helped for the last question.
Matt Kramer Madison, WI
mfkramer@wisc.edu