A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is
![\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B12%5Ccdot%2010%5E9%20Hz%7D%3D0.025%20m)
Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:
![y=\frac{m\lambda D}{a}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bm%5Clambda%20D%7D%7Ba%7D)
where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find
![y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%281%29%280.025%20m%29%2830000%20m%29%7D%7B2.0%20m%7D%3D375%20m)
and so, the diameter is
![d=2y = 750 m](https://tex.z-dn.net/?f=d%3D2y%20%3D%20750%20m)
B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is
![r=\frac{750 m}{2}=375 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B750%20m%7D%7B2%7D%3D375%20m)
So the area is
![A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%28375%20m%29%5E2%3D4.42%5Ccdot%2010%5E5%20m%5E2)
And since the power is
![P=100 kW = 1\cdot 10^5 W](https://tex.z-dn.net/?f=P%3D100%20kW%20%3D%201%5Ccdot%2010%5E5%20W)
The average intensity is
![I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BA%7D%3D%5Cfrac%7B1%5Ccdot%2010%5E5%20W%7D%7B4.42%5Ccdot%2010%5E5%20m%5E2%7D%3D0.23%20W%2Fm%5E2)
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.
</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
We have to convert Gm/s to m/s.
As ![1 \ Gm/s = 10^9 \ m/s](https://tex.z-dn.net/?f=1%20%5C%20Gm%2Fs%20%3D%2010%5E9%20%5C%20m%2Fs)
Therefore the speed of light in vacuum,
![c = 0.3 \ Gm/s = 0.3 \times 10^9 \ m/s \\\\ c= 3 \times 10^8 \ m/s](https://tex.z-dn.net/?f=c%20%3D%200.3%20%5C%20Gm%2Fs%20%3D%200.3%20%5Ctimes%2010%5E9%20%5C%20m%2Fs%20%5C%5C%5C%5C%20c%3D%203%20%5Ctimes%2010%5E8%20%5C%20m%2Fs)
Thus, the speed of light in m/s is ![3 \times 10^8 \ m/s](https://tex.z-dn.net/?f=3%20%5Ctimes%2010%5E8%20%5C%20m%2Fs)