Question

Part A:

An electromagnetic wave is propagating in the positive x direction. At a given moment in time, the magnetic field at the origin points in the positive y direction. In what direction does the electric field at the origin point at that same moment?

Positive x

Negative x

Positive y

Negative y

Positive z

Negative z

Part B:

The figure shows the electromagnetic field as a function of position for two electromagnetic waves traveling in a vacuum at a given moment. Which statement about the frequency and speed of the waves is correct?

The frequency and speed of both waves are equal.

The frequency of wave A is higher and the speed of wave A is greater than the frequency and speed of wave B.

The frequency of wave A is greater than that of wave B, but the speeds of the two waves are the same.

The frequency of wave A is lower and speed of wave A is less than the frequency and speed of wave B.

The frequency of wave A is lower than that of wave B, but the speeds of the two waves are the same. Part C: At a given location in space, the magnetic field in an electromagnetic wave is increasing. How is the electric field changing at that same location? The electric field is decreasing. The electric field is not changing. The electric field is increasing. Part D: Rank the following types of electromagnetic waves by the wavelength of the wave. (Longest to shortest wavelength) Radio Waves, Visible Light, Microwaves, X rays

Answer 1

Concepts and reason

The concepts used in this problem are propagation of electromagnetic wave, electric force, magnetic force and the relation of wavelength with frequency.

First, determine the direction of electric field using the concept of propagation of wave. Then the correct statement for frequency and speed can be determined using the relation of wavelength and frequency. The electric field can be determined using the concept of electric force and magnetic force. Determine the ranking of electromagnetic waves using the wavelength of waves.

Fundamentals

Propagation of electromagnetic wave:

The oscillations of electric field and magnetic field propagate an electromagnetic wave. The magnetic field and electric field are perpendicular to each other and to the direction of propagation. The direction of propagation is given by,

$\hat n = \vec E \times \vec B$

Here, $\hat n$ is the direction of propagation, $\vec E$ is the direction of electric field and $\vec B$ is the direction of magnetic field.

Magnetic force:

The magnetic force is perpendicular to the magnetic field and the velocity of the charge. The magnitude of magnetic force is given by:

$F = qvB$

Here, $q$ is the charge, $v$ is the velocity and $B$ is the magnetic field.

Electric force:

The expression of magnitude of electric force is,

$F = qE$

Here, $q$ is the charge, $E$ is the electric field.

Wavelength and frequency:

The relation between the wavelength and the frequency is,

$f = \frac{v}{\lambda }$

Here, $\lambda$ is the wavelength, $v$ is the speed of wave and $f$ is the frequency of wave.

(Part A)

The direction of propagation of electromagnetic wave is,

$\hat n = \vec E \times \vec B$ …… (1)

The figure drawn below shows the oscillation of electric field and magnetic field and also the propagation of wave.

Substitute $\hat i$ for $\hat n$ and $\hat j$ for $\vec B$ in equation (1).

$\hat i = \vec E \times \hat j$ …… (2)

The cross product identity is.

$- \hat k \times \hat j = \hat i$ …… (3)

From equation (2) and (3),

$\vec E = - \hat k$

The direction is negative z-axis.

(Part B)

The wavelength of wave B is more than the wavelength of wave A.

${\lambda _{\rm{B}}} > {\lambda _{\rm{A}}}$

The frequency is inversely proportional to the wavelength.

The frequency of wave A is,

${f_{\rm{A}}} = \frac{v}{{{\lambda _{\rm{A}}}}}$

The frequency of wave B is,

${f_{\rm{B}}} = \frac{v}{{{\lambda _{\rm{B}}}}}$

$\Rightarrow {f_{\rm{A}}} > {f_{\rm{B}}}$

The speed remains same.

(Part C)

The magnetic force is,

$F = qvB$ …… (4)

The electric force is,

$F = qE$ …… (5)

Equating equation (4) and (5).

$\begin{array}{c}\\qE = qvB\\\\E = vB\\\end{array}$

As the speed is constant, so the electric field is directly proportional to the magnetic field.

The electric field also increases.

(Part D)

The wavelength of X-rays $\left( {{\lambda _{\rm{x}}}} \right)$ ranges from $0.01\;{\rm{nm to 10 nm}}$ .

The wavelength of radio waves $\left( {{\lambda _{\rm{r}}}} \right)$ ranges from $1\;{\rm{mm to 1 km}}$ .

The wavelength range of visible light $\left( {{\lambda _{\rm{v}}}} \right)$ is from $390\;{\rm{nm to 750 nm}}$ .

The wavelength of microwaves ranges $\left( {{\lambda _{\rm{m}}}} \right)$ from $1\;{\rm{mm to 30 cm}}$

$\Rightarrow {\lambda _{\rm{r}}} > {\lambda _{\rm{m}}} > {\lambda _{\rm{v}}} > {\lambda _x}$

Ans: Part A

The direction of electric field is along negative z-direction.

Part B

The frequency of wave A is greater than that of wave B, but the speeds of the two waves are the same.

Part C

The electric field is increasing.

Part D

The ranking of electromagnetic waves from longest to shortest is Radio Wave > Microwave > Visible light > X-ray.