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The electromagnetic power radiated by a nonrelativistic moving point charge \(q\) having an acceleration \(a\) is $$ \mathscr{P}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}} $$ where \(\epsilon_{0}\) is the permittivity of free space and \(c\) is the speed of light in vacuum. (a) Show that the right side of this equation has units of watts. (b) An electron is placed in a constant electric field of magnitude \(100 \mathrm{N} / \mathrm{C}\). Determine the acceleration of the electron and the electromagnetic power radiated by this electron. (c) What If? II a proton is placed in a cyclotron with a radius of \(0.500 \mathrm{m}\) and a magnetic field of magnitude \(0.350 \mathrm{T},\) what electromagnetic power does this proton radiate?

Short Answer

Expert verified
The electromagnetic power radiated by an electron or a proton can be calculated using the equation provided, once the acceleration is found (which varies depending on whether the particle is in an electric field or moving in a cyclotron). Also, the given equation correctly indicates power in watts through dimensional analysis. The exact numerical results depend on the values given or calculated for acceleration, among other parameters.

Step by step solution

01

Dimensional Analysis

To prove the given equation indicates Power in Watts, we need to start by breaking down each variable in the equation into its representational units. \(q\) represents electric charge and is measured in Coulombs (C). \(a\) is the acceleration, measured in meters per second squared (\(m/s^2\)). \(\epsilon_0\) is the permittivity of free space equals approximately \(8.854 \times 10^{-12} F/m\). \(c\) is the speed of light, which equals approximately \(3 \times 10^8 m/s\). Substituting all these into the equation, we get \(C^2 \cdot m^2/s^4\) (for the numerator) and \(F \cdot m^−1 \cdot m^3/s^3\) (for the denominator), equivalently \(C^2 \cdot m^2 \cdot s^2\) for the numerator and \(C^2 \cdot m^2 \cdot s^−3\) for the denominator. The SI unit of power is watt (W), which is equivalent to \(m^2 \cdot kg/s^3\) or \(J/s\). Therefore, the given equation has units of power.
02

Calculate the Acceleration and Radiated Power of Electron

The force acting on the electron due to the electric field is \(F = qE\), where \(E = 100 N/C\) is the electric field. Using Newton's second law, \(F = ma\), one can determine the acceleration of the electron, \(a\). Given that the charge of the electron, \(q = 1.6 \times 10^{-19} C\), and the mass of an electron is \(9.11 x 10^{-31} kg\), the resulting acceleration would be \(a = F/m = qE/m \). The electromagnetic power radiated by this electron is then calculated using the formula \(\mathscr{P}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}}\). Substituting the values appropriately, the radiated power can be calculated.
03

Calculate the Radiated Power of Proton in Cyclotron

In a cyclotron, a charged particle moves in a circle of a given radius due to a perpendicular magnetic field. The magnetic force provides the necessary centripetal force for circular motion. Therefore, \(qvB = mv^2/r\), where \(v\) is the velocity and \(r = 0.5 m\) the radius of the cyclotron. The proton charge \(q = 1.6 x 10^{-19} C\), magnetic field \(B = 0.35 T\), radius \(r = 0.5 m\), and the proton mass \(m = 1.673 x 10^{-27} kg\). From these, we can calculate the velocity as \(v = qrB/m\) and then find the acceleration \(a = v^2/r\). Finally, the radiated power can be calculated using the formula \(\mathscr{P}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}}\) substituting the values calculated.

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Key Concepts

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

Dimensional Analysis in Physics
Dimensional analysis in physics is a technique used to check the consistency of physical equations, converting them into algebra involving units of measurement. It plays a crucial role in determining whether an equation like the electromagnetic power radiated by a charge is dimensionally correct.

In the case of the electromagnetic power radiated, \( \mathscr{P}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}} \), the dimensional analysis involves breaking down the units for charge (Coulombs), acceleration (\(m/s^2\)), the permittivity of free space, and the speed of light to verify that the right side of the equation indeed resolves to Watts, which is the unit of power (\(m^2 \cdot kg/s^3\) or \(J/s\)).

This exercise enforces the student's understanding of the fundamental units of charge, mass, distance, and time, and how they combine to form derived units like power. Understanding dimensional analysis helps students ensure the physical validity of equations before they proceed to numerical calculations.
Acceleration of Charged Particles
When discussing the acceleration of charged particles, it's important to look at how electric and magnetic fields interact with them. In the example of an electron in an electric field, the force exerted on the electron is calculated using \(F = qE\), where \(q\) is the electron charge and \(E\) is the electric field strength.

Using Newton’s second law, \(F=ma\), one can readily find the acceleration by rearranging the formula to \(a = F/m\). For charged particles, their small mass results in a large acceleration for even moderate electric fields, which can then lead to significant electromagnetic power radiated if the particle is accelerating non-relativistically.

This concept is crucial in fields ranging from particle physics to electrical engineering, where the manipulation of charged particles is routine. By calculating the electromagnetic power radiated, which depends on the square of both the charge and the acceleration (\(\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}}\)), students can understand how even tiny particles like electrons can emit measurable amounts of power under the right conditions.
Cyclotron Operation
A cyclotron is an instrument that accelerates charged particles to high speeds using a magnetic field. It operates on the principle that a magnetic field can exert a force on a moving charge, causing it to move in a circular path. In the cyclotron, this path is constrained to a fixed radius, and the particle's speed increases as it spirals outward.

For a proton in a cyclotron with a particular radius and experiencing a magnetic force, the centripetal force needed to maintain its circular path is provided by the magnetic field. The formula \(qvB = mv^2/r\) is pivotal here, as it relates the magnetic field strength (\(B\)), the charge (\(q\)), and the radius of the path (\(r\)) to the velocity (\(v\)) of the proton. From these, students can derive the acceleration and consequently calculate the electromagnetic power radiated using \(\mathscr{P}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}}\).

Understanding cyclotron operation is not only important for solving physics problems but is also essential in applications like particle therapy for cancer treatment, where cyclotrons are used to accelerate protons that can then be directed towards tumors.

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Most popular questions from this chapter

\- Write down expressions for the electric and magnetic fields of a sinusoidal plane electromagnetic wave having a frequency of \(3.00 \mathrm{GHz}\) and traveling in the positive \(x\) direction. The amplitude of the electric field is \(300 \mathrm{V} / \mathrm{m}\).

A very large flat sheet carries a uniformly distributed electric current with current per unit width \(J_{s} .\) Example 30.6 demonstrated that the current creates a magnetic field on both sides of the sheet, parallel to the sheet and perpendicular to the current, with magnitude \(B=\frac{1}{2} \mu_{0} J_{s} .\) If the current oscillates in time according to $$ \mathbf{J}_{s}=J_{\max }(\cos \omega t) \hat{\mathbf{j}}=J_{\max }[\cos (-\omega t)] \hat{\mathbf{j}} $$ the sheet radiates an electromagnetic wave as shown in Figure P34.37. The magnetic field of the wave is described by the wave function \(\quad \mathbf{B}=\frac{1}{2} \mu_{0} J_{\max }[\cos (k x-\omega t)] \hat{\mathbf{k}}\) (a) Find the wave function for the electric field in the wave. (b) Find the Poynting vector as a function of \(x\) and \(t\) (c) Find the intensity of the wave. (d) What If? If the sheet is to emit radiation in each direction (normal to the plane of the sheet) with intensity \(570 \mathrm{W} / \mathrm{m}^{2},\) what maximum value of sinusoidal current density is required?

A linearly polarized microwave of wavelength \(1.50 \mathrm{cm}\) is directed along the positive \(x\) axis. The electric field vector has a maximum value of \(175 \mathrm{V} / \mathrm{m}\) and vibrates in the \(x y\) plane. (a) Assume that the magnetic field component of the wave can be written in the form \(B=B_{\max } \sin (k x-\omega t)\) and give values for \(B_{\max }, k,\) and \(\omega\) Also, determine in which plane the magnetic field vector vibrates. (b) Calculate the average value of the Poynting vector for this wave. (c) What radiation pressure would this wave exert if it were directed at normal incidence onto a perfectly reflecting sheet? (d) What acceleration would be imparted to a \(500-\mathrm{g}\) sheet (perfectly reflecting and at normal incidence) with dimensions of \(1.00 \mathrm{m} \times 0.750 \mathrm{m} ?\)

An electromagnetic wave in vacuum has an electric field amplitude of \(220 \mathrm{V} / \mathrm{m} .\) Calculate the amplitude of the corresponding magnetic field.

Consider a small, spherical particle of radius \(r\) located in space a distance \(R\) from the Sun. (a) Show that the ratio \(F_{\mathrm{rad}} / F_{\mathrm{grav}}\) is proportional to \(1 / r,\) where \(F_{\mathrm{rad}}\) is the force exerted by solar radiation and \(F_{\mathrm{grav}}\) is the force of gravitational attraction. (b) The result of part (a) means that, for a sufficiently small value of \(r,\) the force exerted on the particle by solar radiation exceeds the force of gravitational attraction. Calculate the value of \(r\) for which the particle is in equilibrium under the two forces. (Assume that the particle has a perfectly absorbing surface and a mass density of \(1.50 \mathrm{g} / \mathrm{cm}^{3} .\) Let the particle be located \(3.75 \times 10^{11} \mathrm{m}\) from the Sun, and use \(214 \mathrm{W} / \mathrm{m}^{2}\) as the value of the solar intensity at that point.)

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