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High-Energy Cancer Treatment. Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012W) pulses of light 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 mm in diameter, with the pulse lasting for 4.0 ns with an average power of 2.0×1012W. 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 ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

Short Answer

Expert verified

a. 8 kJ of energy is given to the cell during pulse.

b. The amount of intensity delivered to cell is 1021W/m2.

c. The maximum values of the electric and magnetic fields in the pulse are 8.7×1011V/mand 2.9×103Trespectively.

Step by step solution

01

Define the intensity ( I  ) and define the formulas.

The power transported per unit area is known as the intensity ( I ) .

The formula used to calculate the intensity ( I ) is:

I=PA

Where, Ais area measured in the direction perpendicular to the energy andP is the power in watts.

The formula used to determine the amplitude of electric and magnetic fields of the wave are:

Emax=2Iε0cBmax=Emaxc

Where,ε0=8.85×10-12C2/N·m2 and c is the speed of light that is equal to 3.0×108m/s.

The energy of flows through the area is related to time as F=Pt.

02

Determine the energy.

Given that,

P=2×1012Wt=4×10-9s

The energy of flows through the area is related to time is

F=Pt

Substitute values in above equation

F=2×10124×10-9=8kJ

Hence,8kJ of energy is given to the cell during pulse.

03

Determine the intensity.

Given that,

P=2×1012Wd=5μm

The formula used to calculate the intensityI is:

I=PA

Substitute the values

I=2×1012π5×10-62=2×101219.63×10-12=1021W/m2

Hence, the amount of intensity delivered to cell is 1021W/m2.

04

Determine the maximum values of electric and magnetic fields.

The maximum value of electric field is:

Emax=2Iε0c

Substitute the values

Emax=2×10218.85×10-123×108=8.7×1011V/m

The amplitude of magnetic field is:

Bmax=Emaxc

Substitute the values

Bmax=8.7×10113×108=2.9×103T

Hence, the maximum values of the electric and magnetic fields in the pulse are8.7×1011V/m and2.9×103T respectively.

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