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

Short Answer

Expert verified
(a) 80 J, (b) \(1.02 \times 10^{23}\) W/m extsuperscript{2}, (c) \(E_0 = 8.69 \times 10^{11}\) V/m, \(B_0 = 2.90 \times 10^{3}\) T.

Step by step solution

01

Calculate Total Energy of the Pulse

The total energy of the pulse can be calculated using the formula \( E = P \times t \), where \( P = 2.0 \times 10^{12} \) Watts is the power and \( t = 4.0 \) nanoseconds \( = 4.0 \times 10^{-9} \) seconds. Calculate \( E = (2.0 \times 10^{12}) \times (4.0 \times 10^{-9}) = 8.0 \times 10^{3} \) Joules.
02

Energy Given to One Cell

Since the pulse energy is spread over 100 cells, the energy delivered to one cell is \( \frac{8.0 \times 10^{3}}{100} = 80 \) Joules.
03

Calculate Intensity Delivered to the Cell

First, find the area of one cell, modeled as a circular disk with diameter \( 5.0 \mu m = 5.0 \times 10^{-6} \) meters. The radius \( r \) is half the diameter: \( r = 2.5 \times 10^{-6} \) m. Use the formula for the area of a circle, \( A = \pi r^2 = \pi (2.5 \times 10^{-6})^2 \). Calculate \( A = \pi \times 6.25 \times 10^{-12} \approx 1.96 \times 10^{-11} \text{ m}^2 \). The intensity \( I = \frac{P}{A} = \frac{2.0 \times 10^{12}}{1.96 \times 10^{-11}} \), resulting in \( I \approx 1.02 \times 10^{23} \text{ W/m}^2 \).
04

Calculate Maximum Values of Electric and Magnetic Fields

For electromagnetic waves, the intensity is related to the maximum electric field \( E_{0} \) by \( I = \frac{1}{2} \varepsilon_0 c E_0^2 \). Solving for \( E_0 \), we have \( E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} \). Using \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \) and \( c = 3.00 \times 10^8 \text{ m/s} \), calculate \( E_0 = \sqrt{\frac{2 \times 1.02 \times 10^{23}}{8.85 \times 10^{-12} \times 3.00 \times 10^8}} \approx 8.69 \times 10^{11} \text{ V/m} \). To find the magnetic field \( B_0 \), use \( B_0 = \frac{E_0}{c} \), yielding \( B_0 \approx \frac{8.69 \times 10^{11}}{3.00 \times 10^8} \approx 2.90 \times 10^{3} \text{ T} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cancer Cell Treatment
The innovative approach to cancer treatment involves using ultrahigh-energy light pulses. These pulses deliver immense power to target cancer cells precisely. Unlike traditional methods that risk damaging surrounding tissues, this approach carefully directs energy to the cells, ensuring targeted disruption.

These light pulses can effectively scramble the internal machinery of a cancer cell within nanoseconds. The key advantage is not causing the cell to explode, thus preventing collateral damage. This ability to target cells so efficiently holds promise in improving cancer therapy.

By precisely controlling the duration and intensity of the pulses, scientists aim to develop treatments that minimize side effects and enhance patient recovery.
Electromagnetic Waves
Electromagnetic waves, which include light pulses, are central to this technique. These waves are oscillations of electric and magnetic fields that travel through space, transporting energy. Light pulses used in cancer treatments are a specific type of electromagnetic wave.

They have unique properties that allow them to interact with biological tissues at a cellular level. The waves' high frequency and short wavelength enable them to penetrate and affect cells precisely In this application, carefully calculated light pulses are calibrated to maximize their interaction with cancerous cells while avoiding harming healthy tissues.

Understanding electromagnetic waves' nature is crucial in harnessing their potential for medical treatments, particularly in targeting and disrupting cancer cells.
Energy Calculation
The process of calculating the energy delivered by light pulses involves basic physics principles. The formula used is:
  • The power of the pulse is multiplied by the time duration it lasts: \( E = P \times t \).
  • Having a power of \(2.0 \times 10^{12}\) Watts and a pulse duration of \(4.0\) nanoseconds \((4.0 \times 10^{-9}\sec)\), we determine the total energy: \( E = 8.0 \times 10^{3} \) Joules.
This energy is dispersed across 100 cells, so each cell receives 80 Joules.

This calculation is vital in ensuring precise energy delivery to target cells, preventing both under-treatment and potential damage to surrounding healthy tissues.
Intensity Calculation
Determining the intensity of the light pulse on a cell's surface is a crucial step in the treatment's effectiveness. The intensity is defined as the power per unit area delivered to a surface:

  • We start by calculating the cell's area, modeled as a circular disk with a diameter of \(5.0 \mu m\), or \(5.0 \times 10^{-6}\) meters.
  • Using the area of a circle, \( A = \pi r^2 \), where the radius \( r = 2.5 \times 10^{-6} \) m, the area becomes \( A \approx 1.96 \times 10^{-11} \, \text{m}^2 \).
  • The intensity \( I \) is then calculated as \( \frac{P}{A} \), yielding \( I \approx 1.02 \times 10^{23} \text{ W/m}^2 \).
Intensity gives insight into how concentrated the energy is, ensuring sufficient disruption of cancerous cells while maintaining precision.
Electric and Magnetic Fields
Understanding electric and magnetic fields is fundamental when dealing with electromagnetic waves. In cancer treatments using light pulses:

  • The intensity of the wave is related to the electric field's maximum value \( E_{0} \) through the equation \( I = \frac{1}{2} \varepsilon_0 c E_0^2 \).
  • Using known constants: \( \varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2) \) and \( c \approx 3.00 \times 10^8 \text{ m/s} \), we calculate \( E_0 \approx 8.69 \times 10^{11} \text{ V/m} \).
  • The magnetic field \( B_0 \) is derived from the electric field: \( B_0 = \frac{E_0}{c} \), which results in \( B_0 \approx 2.90 \times 10^{3} \text{ T} \).
These calculations are essential in analyzing how the fields interact with the cancer cells, determining the pulse's effectiveness and safety for the surrounding tissues.

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Most popular questions from this chapter

The intensity of a cylindrical laser beam is 0.800 W/m\(^2\). The cross- sectional area of the beam is 3.0 \(\times\) 10\(^{-4}\) m\(^2\) and the intensity is uniform across the cross section of the beam. (a) What is the average power output of the laser? (b) What is the rms value of the electric field in the beam?

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electricfield vector \(\vec{E}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (\(Hint\): Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

The energy flow to the earth from sunlight is about 1.4 kW/m\(^2\). (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about 1.5 \(\times\) 10\(^{11}\) m. Find the total power radiated by the sun.

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the +\(x\)-direction, \(\vec{B}\) in the +\(y\)-direction; (b) \(\vec{E}\) in the -\(y\)-direction, \(\vec{B}\) in the +\(x\)-direction; (c) \(\vec{E}\) in the +\(z\)-direction, \(\vec{B}\) in the -\(x\)-direction; (d) \(\vec{E}\) in the +\(y\)-direction, \(\vec{B}\) in the -\(z\)-direction.

Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm in air. What are the frequency, period, and wave number of such waves?

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