Scientists are working on a new technique to kill cancer...

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahighenergy (in the range of $10^{12} \mathrm{~W}$ ) 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 $5.0 \mu \mathrm{m}$ in diameter, with the pulse lasting for $4.0 \mathrm{~ns}$ with an average power of $2.0 \times 10^{12} \mathrm{~W}$. 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? (b) What is the intensity (in $\mathrm{W} / \mathrm{m}^{2}$ ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

Transcript

00:01 Hi there.

00:02 So for this problem, we are told that scientists are working on a new technique to kill cancer cells by sapping them with ultra -high energy pulses of electromagnetic waves that lasts for an extremely short time.

00:23 And that is a few nanoseconds.

00:26 So these short pulses scramble the interior of a cell without a short.

00:31 Causing it to explode, as long pulses will do.

00:36 We can model typically such a cell as a disk with a diameter that is given, and the diameter is equal to five micrometers.

00:47 Now remember that micro means 10 to the minus 6 meters.

00:53 And with the pull lasting for a time, so let's set that this is the times t, that is equal to four nanoseconds.

01:03 That is the same as four times 10 to the minus nine seconds.

01:08 And with an average power that is also given, and that is equal to two times 10 to the 12 watts.

01:23 Now, we shall assume that the energy is spread uniformly over the phases of a 100 cells for each.

01:34 Pools.

01:35 So for part a of this problem, we are asked about how much energy is given to the cell during these pools.

01:47 Now, recall that the power is defined as the energy divided by the time.

01:55 So we need to solve for the energy in this case.

01:59 So that will be the power times the time.

02:02 So that will be that the energy is equal to the power that we are given, 2 times 10 to the 12 whaps, this times the time, 4 times 10 to the minus 9 seconds that we are given, which is the same as 4 nanoseconds.

02:20 So the energy that we obtain from this is 8 times 10 to the 3 joules, which is the same as 8 ,000 joules.

02:30 Now, this is the total energy.

02:32 Which is spread out over 100 cells.

02:35 So the energy for each individual cell is just, so the energy for each individual, that cell is just simply 80 joules.

02:47 So that's a solution for part a of this problem.

02:52 Now for part b, we are asked about what is the intensity delivered to the cell? so in this part of the problem, in order to obtain the intensity, we need to divide the power by the area.

03:09 So the area for the cell is just the area of a circle.

03:14 So the area in this case is just pi times its radius square.

03:18 Now remember that in this case we are given the diameter.

03:22 So we can write this as pi times the diameter squared divided by four because remember that the radius is defined as the diameter divided by two.

03:33 So now we substitute the values in here.

03:35 So that will be pi times the diameter that is 5 times 10 to the minus 6 meters.

03:40 That to the square this divided by 4.

03:42 So the area that we obtain from this is equal to 2 times 10 to the minus 11 meters square.

03:50 Now we need to divide the power by this area.

03:54 So the intensity is just equal to the power divided by 100 times the area because the area that we just obtained is the area for one cell...