/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 BI0 High-Energy Cancer Treatment... [FREE SOLUTION] | 91Ó°ÊÓ

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BI0 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 \(10^{12} \mathrm{~W}\) ) 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 \(5.0 \mu \mathrm{m}\) 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?

Short Answer

Expert verified
(a) 0.08 J, (b) 1.02 x 10^14 W/m^2, (c) Max electric field = 8.65 x 10^7 V/m, Max magnetic field = 0.29 T.

Step by step solution

01

Calculate Total Energy of Pulse

The energy provided by the pulse can be obtained using the formula \[ E = P \times t \]where \( E \) is the energy, \( P = 2.0 \times 10^{12} \text{ W} \) is the power, and \( t = 4.0 \text{ ns} = 4.0 \times 10^{-9} \text{ s} \) is the time duration of the pulse. Thus,\[E = (2.0 \times 10^{12} \text{ W})(4.0 \times 10^{-9} \text{ s}) = 8.0 \text{ J}.\]
02

Calculate Energy Delivered per Cell

As the energy is distributed equally over 100 cells, the energy per cell is:\[E_{\text{cell}} = \frac{E}{100} = \frac{8.0 \text{ J}}{100} = 0.08 \text{ J}.\]
03

Calculate Area of a Cell

The area \( A \) of one cell is calculated using the formula for the area of a circle:\[ A = \pi \left(\frac{d}{2}\right)^2 \]where \( d = 5.0 \mu \text{m} = 5.0 \times 10^{-6} \text{ m} \). Substituting these values:\[A = \pi \left(\frac{5.0 \times 10^{-6} \text{ m}}{2}\right)^2 \approx 1.9635 \times 10^{-11} \text{ m}^2.\]
04

Calculate Intensity Delivered to the Cell

Intensity \( I \) is the power delivered per unit area, given by:\[ I = \frac{P}{A} \]Substituting the power for the cell and the area calculated in previous steps:\[ I = \frac{2.0 \times 10^{12} \text{ W}}{1.9635 \times 10^{-11} \text{ m}^2 \times 100} \approx 1.02 \times 10^{14} \text{ W/m}^2.\]
05

Calculate Maximum Electric Field

The intensity \( I \) and the electric field \( E \) are related by the equation:\[ I = \frac{1}{2} c \varepsilon_0 E^2 \]where \( c = 3.0 \times 10^8 \text{ m/s} \) is the speed of light and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of vacuum. Solving for \( E \),\[ E = \sqrt{\frac{2I}{c \varepsilon_0}} = \sqrt{\frac{2 \times 1.02\times 10^{14}}{3.0 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 8.65 \times 10^7 \text{ V/m}.\]
06

Calculate Maximum Magnetic Field

The magnetic field \( B \) is related to the electric field \( E \) by:\[ B = \frac{E}{c} \]Substituting the value of \( E \),\[ B = \frac{8.65 \times 10^7 \text{ V/m}}{3.0 \times 10^8 \text{ m/s}} \approx 0.29 \text{ T}.\]

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Key Concepts

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

Cancer Treatment
High-energy physics is playing a crucial role in advancing cancer treatment techniques. The primary aim of these newly developed methods is to effectively target and destroy cancerous cells while minimizing damage to the surrounding healthy tissues. This innovative approach employs ultrahigh-energy pulses of light to tackle the cancer cells directly.

By utilizing these energetic light pulses, which have the power of around \(10^{12} \) watts, scientists can disrupt the internal structure of the cancer cells. These pulses are incredibly brief, typically lasting just a few nanoseconds. This brief duration allows the pulse to scatter the internal components of a cell without causing it to explode, which is a crucial distinction when compared to using longer energy exposures.

Through careful design, these techniques concentrate their effects precisely on the malignant cells, thereby providing a new angle in powerful and focused cancer therapy solutions.
Ultrahigh-Energy Pulses
Ultrahigh-energy pulses are a fascinating topic in the field of high-energy physics, with promising applications beyond traditional boundaries. These pulses are characterized by their enormous power delivered over an exceptionally brief duration.

Such pulses are measured in terawatts (\(10^{12} ext{ W}\)), indicating a vast amount of energy delivered in an instant. In cancer treatment applications, these pulses are precisely targeted and generated to last only a few nanoseconds, ensuring they focus solely on their intended target and minimize collateral damage. The short duration is crucial as it maintains the integrity of the surrounding tissue while effectively scrambling the targeted cancer cell's interior.

This method is part of cutting-edge research that opens up new possibilities not only for medical applications but also for advanced technological innovations where brief, intensely focused energy delivery is required.
Electric Field Calculations
Electric field calculations are central to understanding how ultrahigh-energy pulses interact with cancer cells during treatment. The intensity of the energy delivered by the pulse is represented as power per unit area and is essential for assessing the effectiveness of the treatment.

To find the maximum electric field within the pulse, we use the formula related to the intensity \(I\) given by:\[ I = \frac{1}{2} c \varepsilon_0 E^2 \]
where \(c\) is the speed of light and \(\varepsilon_0\) is the permittivity of the vacuum. Solving for the electric field \(E\), provides crucial insights into the force exerted by the pulse.

Understanding these calculations allows researchers to optimize the pulse characteristics, ensuring maximum efficiency and safety in disrupting cancer cells without harming the surrounding tissues. It is a sophisticated balance of physics and biology at work.
Magnetic Field Calculations
The relationship between electric and magnetic fields is fundamental in contexts like high-energy cancer treatment. Knowing the electric field helps us determine the magnetic field value, which is also influenced by the pulse.

The connection is given by the formula:\[ B = \frac{E}{c} \]
where \(B\) is the magnetic field, \(E\) is the electric field calculated earlier, and \(c\) is the speed of light. This relationship comes from the nature of electromagnetic waves, where electric and magnetic fields support each other as the wave propagates.

Ensuring accurate calculations of both fields is vital for designing medical treatments that require precise energy delivery. By mastering these principles, scientists can better control how these ultrahigh-energy pulses affect cancer cells, paving the way for more effective and less invasive protocols.

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Most popular questions from this chapter

A satellite \(575 \mathrm{~km}\) above the earth's surface transmits sinusoidal electromagnetic waves of frequency \(92.4 \mathrm{MHz}\) uniformly in all directions, with a power of \(25.0 \mathrm{~kW}\). (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring \(15.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm}\) oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

Flashlight to the Rescue. You are the sole crew member of the interplanetary spaceship \(T: 1339\) Vorga, which makes regular cargo runs between the earth and the mining colonies in the asteroid belt. You are working outside the ship one day while at a distance of \(2.0 \mathrm{AU}\) from the sun. [1 AU (astronomical unit) is the average distance from the earth to the sun, \(149,600,000 \mathrm{~km} .]\) Unfortunately, you lose contact with the ship's hull and begin to drift away into space. You use your spacesuit's rockets to try to push yourself back toward the ship, but they run out of fuel and stop working before you can return to the ship. You find yourself in an awkward position, floating \(16.0 \mathrm{~m}\) from the spaceship with zero velocity relative to it. Fortunately, you are carrying a 200-W flashlight. You turn on the flashlight and use its beam as a "light rocket" to push yourself back toward the ship. (a) If you, your spacesuit, and the flashlight have a combined mass of \(150 \mathrm{~kg}\), how long will it take you to get back to the ship? (b) Is there another way you could use the flashlight to accomplish the same job of returning you to the ship?

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 electric-field vector \(\overrightarrow{\boldsymbol{E}}\) 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 \(\overrightarrow{\boldsymbol{B}}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\overrightarrow{\boldsymbol{S}}\) 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 currentcarrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

An electromagnetic wave has an electric field given by \(\overrightarrow{\boldsymbol{E}}(y, t)=\left(3.10 \times 10^{5} \mathrm{~V} / \mathrm{m}\right) \hat{k} \cos \left[\mathrm{ky}-\left(12.65 \times 10^{12} \mathrm{rad} / \mathrm{s}\right) t\right]\) (a) In which direction is the wave traveling? (b) What is the wavelength of the wave? (c) Write the vector equation for \(\vec{B}(y, t)\)

An electromagnetic wave with frequency \(65.0 \mathrm{~Hz}\) travels in an insulating magnetic material that has dielectric constant \(3.64\) and relative permeability \(5.18\) at this frequency. The electric field has amplitude \(7.20 \times 10^{-3} \mathrm{~V} / \mathrm{m}\). (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

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