Question

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of electromagnetic waves that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 4.0 μm in diameter, with the pulse lasting for 5.0 ns with an average power of 1.59×1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.a. How much energy is given to the cell during this pulse? U=___Jb. What is the intensity (in W/m^2) delivered tothe cell? I=___W/m^2c. What is the maximum value of the electric field in the pulse?Emax=___V/md. What is the maximum value of the magnetic field in the pulse?Bmax=___T

Answer 1

1. $7.95\cdot 10^6 J$

The total energy given to the cells during one pulse is given by:

$E=Pt$

where

P is the average power of the pulse

t is the duration of the pulse

In this problem,

$P=1.59\cdot 10^{12}W$

$t=5.0 ns = 5.0\cdot 10^{-9} s$

Substituting,

$E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J$

2. $1.26\cdot 10^{21}W/m^2$

The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is

$r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m$

So the area of each cell is

$A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2$

The energy is spread over 100 cells, so the total area of the cells is

$A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2$

And so the intensity delivered is

$I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2$

3. $9.74\cdot 10^{11} V/m$

The average intensity of an electromagnetic wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

E is the amplitude of the electric field

Solving the formula for E, we find:

$E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m$

4. 3247 T

The magnetic field amplitude is related to the electric field amplitude by

$E=cB$

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

$B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T$