1. ![7.95\cdot 10^6 J](https://tex.z-dn.net/?f=7.95%5Ccdot%2010%5E6%20J)
The total energy given to the cells during one pulse is given by:
![E=Pt](https://tex.z-dn.net/?f=E%3DPt)
where
P is the average power of the pulse
t is the duration of the pulse
In this problem,
![P=1.59\cdot 10^{12}W](https://tex.z-dn.net/?f=P%3D1.59%5Ccdot%2010%5E%7B12%7DW)
![t=5.0 ns = 5.0\cdot 10^{-9} s](https://tex.z-dn.net/?f=t%3D5.0%20ns%20%3D%205.0%5Ccdot%2010%5E%7B-9%7D%20s)
Substituting,
![E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J](https://tex.z-dn.net/?f=E%3D%281.59%5Ccdot%2010%5E%7B12%7DW%29%285.0%20%5Ccdot%2010%5E%7B-6%7Ds%29%3D7.95%5Ccdot%2010%5E6%20J)
2. ![1.26\cdot 10^{21}W/m^2](https://tex.z-dn.net/?f=1.26%5Ccdot%2010%5E%7B21%7DW%2Fm%5E2)
The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is
![r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B4.0%5Cmu%20m%7D%7B2%7D%3D2.0%20%5Cmu%20m%20%3D%202.0%5Ccdot%2010%5E%7B-6%7Dm)
So the area of each cell is
![A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%282.0%20%5Ccdot%2010%5E%7B-6%7Dm%29%5E2%3D1.26%5Ccdot%2010%5E%7B-11%7D%20m%5E2)
The energy is spread over 100 cells, so the total area of the cells is
![A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2](https://tex.z-dn.net/?f=A%3D100%20%281.26%5Ccdot%2010%5E%7B-11%7D%20m%5E2%29%3D1.26%5Ccdot%2010%5E%7B-9%7D%20m%5E2)
And so the intensity delivered is
![I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BA%7D%3D%5Cfrac%7B1.59%5Ccdot%2010%5E%7B12%7DW%7D%7B1.26%5Ccdot%2010%5E%7B-9%7D%20m%5E2%7D%3D1.26%5Ccdot%2010%5E%7B21%7DW%2Fm%5E2)
3. ![9.74\cdot 10^{11} V/m](https://tex.z-dn.net/?f=9.74%5Ccdot%2010%5E%7B11%7D%20V%2Fm)
The average intensity of an electromagnetic wave is related to the maximum value of the electric field by
![I=\frac{1}{2}c\epsilon_0 E^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7B2%7Dc%5Cepsilon_0%20E%5E2)
where
c is the speed of light
is the vacuum permittivity
E is the amplitude of the electric field
Solving the formula for E, we find:
![E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m](https://tex.z-dn.net/?f=E%3D%5Csqrt%7B%5Cfrac%7B2I%7D%7Bc%5Cepsilon_0%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%281.26%5Ccdot%2010%5E%7B21%7D%20W%2Fm%5E2%29%7D%7B%283%5Ccdot%2010%5E8%20m%2Fs%29%288.85%5Ccdot%2010%5E%7B-12%7DF%2Fm%29%7D%7D%3D9.74%5Ccdot%2010%5E%7B11%7D%20V%2Fm)
4. 3247 T
The magnetic field amplitude is related to the electric field amplitude by
![E=cB](https://tex.z-dn.net/?f=E%3DcB)
where
E is the electric field amplitude
c is the speed of light
B is the magnetic field
Solving the equation for B and substituting the value of E that we found at point 3, we find
![B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7BE%7D%7Bc%7D%3D%5Cfrac%7B9.74%5Ccdot%2010%5E%7B11%7D%20V%2Fm%7D%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%3D3247%20T)