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A sinusoidal electromagnetic wave is propagating in vacuum in the \(+z\) -direction. If at a particular instant and at a certain point in space the electric field is in the \(+x\) -direction and has magnitude \(4.00 \mathrm{~V} / \mathrm{m}\) what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

Short Answer

Expert verified
The magnetic field of the wave at this same point in space and instant in time is \(1.33 \times 10^{-8} \, T\) and is in the \('+y'\) direction.

Step by step solution

01

Identify Given Parameters

The direction of propagation of the electromagnetic wave is given as \('+z\'). The electric field \('E'\) is in the \('+x\') direction with a magnitude of \(4.00 \mathrm{~V/m}\). The goal is to determine the magnitude and direction of the magnetic field \('B'\) at this same point in space and the same instant in time.
02

Apply the Relationship

According to Maxwell's equations, the magnitude of the electric field \('E'\) and the magnetic field \('B'\) are related in an electromagnetic wave by the formula \(E = cB\), where \(c\) is the speed of light in a vacuum. We can rearrange the formula to solve for \(B\): \(B = E / c\).
03

Calculate the Magnetic Field

Using the values for the electric field \(E = 4.00V/m\) and the speed of light \(c = 3.00 \times 10^{8} m/s\), we insert them into the equation found in the previous step to calculate the magnetic field: \(B = (4.00 \, V/m) / (3.00 \times 10^{8} \, m/s) = 1.33 \times 10^{-8} \, T\). The units of \(B\) are Tesla, represented by 'T', which corresponds to \(\, V/(m \cdot s)\).
04

Determine the Direction of the Magnetic Field

Since the magnetic field, electric field, and direction of propagation are all perpendicular to each other and follow the right-hand rule, and since we know the electric field is in the \('+x'\) direction and the direction of propagation is \('+z'\), we can deduce that the magnetic field \('B'\) is in the \('+y'\) direction using the right-hand rule.

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Key Concepts

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

Maxwell's Equations
Maxwell's equations are a set of four fundamental laws that describe the behavior of electric and magnetic fields. These equations form the foundation for understanding how electromagnetic waves, such as light, propagate through space. Here is a summary of each law:

  • **Gauss's Law for Electricity** states that the electric flux through a closed surface is proportional to the charge enclosed. Mathematically, it is expressed as \( abla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}\).
  • **Gauss's Law for Magnetism** states that there are no magnetic monopoles; the total magnetic flux out of a closed surface is zero, written as \( abla \cdot \mathbf{B} = 0\).
  • **Faraday's Law of Induction** highlights how a changing magnetic field can induce an electric field. It is represented by \( abla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}\).
  • **Ampère-Maxwell Law** extends Ampère's Law by including a term for changing electric fields, represented by \( abla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\).
These equations collectively describe how electromagnetic fields are generated and altered by each other and by charges and currents. In the context of electromagnetic waves, they indicate how electric and magnetic fields propagate through space and are interlinked.
Electric Field
The electric field is a fundamental concept in electromagnetism. It is a vector field that represents the force exerted on charged particles. The electric field is produced by charged objects or by a changing magnetic field, as per Faraday's Law.

In the context of an electromagnetic wave, the electric field oscillates perpendicular to the direction of wave propagation. For the given exercise, the electric field is directed along the \( +x\) axis while the wave travels along the \(+z\) axis. This perpendicular relationship is a standard characteristic of electromagnetic waves.
  • The strength or magnitude of the electric field at a point is expressed in volts per meter (V/m).
  • The direction of the electric field vector indicates the direction of the force that a positive test charge would experience.
  • In vacuum, the electric field propagates at the speed of light without attenuation.
Understanding electric fields is essential because it allows us to predict the behavior of charged particles in various configurations.
Magnetic Field
A magnetic field is another pivotal vector field essential in the study of electromagnetism. In an electromagnetic wave, this field is also an oscillating component that is interdependent with the electric field.

The magnetic field oscillates perpendicular to both the electric field and the direction of wave propagation. In the exercise example, while the wave travels in the \( +z\) direction and the electric field points in the \( +x\) direction, the magnetic field must align along the \( +y\) axis to maintain this perpendicular arrangement.
  • The units of magnetic field strength are Tesla (T), with an alternative in V/(m·s).
  • Magnetic fields are generated by moving electric charges or changing electric fields, as described by Maxwell's equations.
  • Unlike electric fields, magnetic fields do not have start or end points but rather form closed loops.
The mutual creation and propagation of the electric and magnetic fields at the speed of light underpin the very nature of electromagnetic radiation.
Right-Hand Rule
The right-hand rule is a useful mnemonic for determining the direction of vectors in electromagnetic fields, particularly within the context of electromagnetic waves. It helps determine the orientation of the electric field, magnetic field, and the wave's propagation direction.

To use the right-hand rule in electromagnetic contexts, follow these simple steps:
  • Point your thumb in the direction of wave propagation, which, in this exercise, is \( +z\).
  • Extend your index finger in the direction of the electric field, or \( +x\).
  • Your middle finger, when held perpendicular to your thumb and index finger, will point in the direction of the magnetic field, which is \( +y\) in this case.
By utilizing the right-hand rule, you can ensure that the relationship between the electric and magnetic fields and the direction of wave travel are all consistent and correct. This rule is invaluable when working with both theoretical and practical applications of electromagnetism.

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Most popular questions from this chapter

We can reasonably model a \(75 \mathrm{~W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\left.\mathrm{W} / \mathrm{m}^{2}\right)\) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

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, \(\overrightarrow{\boldsymbol{B}}\) in the \(+y\) -direction; (b) \(\overrightarrow{\boldsymbol{E}}\) in the \(-y\) -direction, \(\vec{B}\) in the \(+x\) -direction; (c) \(\vec{E}\) in the \(+z\) -direction, \(\vec{B}\) in the \(-x\) -direction; (d) \(\overrightarrow{\boldsymbol{E}}\) in the \(+y\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-z\) -direction.

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from \(\mathrm{WCCO}\) is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

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