Electromagnetic Waves

From Maxwell’s equations, in electromagnetic wave:

Radiation pressure on black body:

Radiation pressure at distance r:

The formula for self inductance: Given:

Force on a reflective surface:

The electric and magnetic fields in an electromagnetic wave are related by: Given: Frequency: Angular frequency: Wave number: Magnetic field amplitude: Since wave propagates in x-direction and electric field in y-direction, magnetic field is along z-direction:

Magnetic field in solenoid: Change in flux:

In electromagnetic waves:

For efficient transmission and reception of electromagnetic waves, the size of the antenna should generally be comparable to the wavelength ( ) of the waves being transmitted or received. A common rule of thumb for a minimum or practical size, especially for a simple dipole antenna, is that its length should be of the order of or . The question asks for the "minimum size," which often relates to . First, calculate the wavelength ( ) of the electromagnetic waves. The relationship between wavelength ( ), frequency ( ), and the speed of light ( ) is . So, . Given values: Frequency . Speed of light . Calculate : . . . So, . Convert wavelength to centimeters: . Now, consider the minimum size of the antenna. A common practical minimum length for an efficient antenna (like a quarter-wave monopole or one arm of a half-wave dipole) is . Minimum size . Minimum size . This matches option (b). If the antenna were a half-wave dipole, its total length would be . The term "size" could refer to the characteristic dimension. For many antenna types, is a key dimension. Given the options, 5 cm is a strong candidate.

In a plane electromagnetic wave propagating in vacuum or a non-dispersive medium, the energy is equally shared between the electric field and the magnetic field. The instantaneous energy density of the electric field ( ) is given by , where is the instantaneous electric field strength and is the permittivity of free space. The instantaneous energy density of the magnetic field ( ) is given by , where is the instantaneous magnetic field strength and is the permeability of free space. For an electromagnetic wave, the magnitudes of the electric and magnetic fields are related by , where is the speed of light in vacuum. Also, . Substitute into the expression for : . Since , . This is exactly the expression for . So, for the instantaneous energy densities. If and are the average energy densities, they are obtained by averaging the instantaneous values over one period or a long time. Since at every instant, their averages must also be equal. . . Since at all times, it follows that .

In an electromagnetic wave: Given: - - Since , and propagation is along , then

In an electromagnetic wave traveling in vacuum or free space:
- Electric and magnetic fields are perpendicular and in phase.
- Energy densities in both fields are equal:
And,

Average power per unit area (Intensity):

The average energy density of an electromagnetic wave can be found using the formula for energy density, which is proportional to the square of the electric field. From the given electric field equation, we calculate the energy density. Thus, the correct answer is option (2).

For an electromagnetic wave in vacuum, the relation between and is: Substitute into the first equation: Thus, the correct relationship is given in option (3).

Useful power emitted as radiation = The intensity at : Peak electric field: Using simplified relation for :

In an electromagnetic wave, the electric field vector and the magnetic field vector are mutually perpendicular to each other and are both perpendicular to the direction of propagation of the wave.
The direction of propagation of the electromagnetic wave is given by the direction of the Poynting vector , which represents the energy flux density (power per unit area) of the electromagnetic field.
The Poynting vector is defined as: $ $ where is the permeability of free space.
The direction of the Poynting vector is the same as the direction of the cross product .
The cross product of two vectors results in a vector that is perpendicular to both original vectors, and its direction is given by the right-hand rule.
In the case of an electromagnetic wave, if and are perpendicular, their cross product points in the direction of the wave's propagation.
Therefore, the direction of propagation of the electromagnetic wave is parallel to the direction of .

The unit of permeability, , in a magnetic context is given by , which is equivalent to . - Option (1), , is a correct unit of permeability. - Option (2), , is also a correct unit of permeability. - Option (3), , is another valid representation of permeability. However, option (4), , does not represent permeability as it is related to electric fields, not magnetic permeability. Thus, the correct answer is option (4). % Final Answer The correct answer is .

The frequency of the EM wave produced is the same as the frequency of charge oscillation.

We need to find the momentum delivered by electromagnetic waves to a surface.
Step 1: Calculate the energy delivered.
Energy flux: Area: Power: Energy in 1 second: Step 2: Relate energy to momentum.
For complete absorption: Final Answer:

The energy density of an electromagnetic wave is proportional to the square of the electric field. If the electric field increases by 100%, the energy density increases by , or 400%.

Step 1: Apply Energy Density Formula The average energy density is given by: Step 2: Compute Energy Density Thus, the correct answer is J/m³.

The intensity of an electromagnetic wave is related to the rms value of the electric field by: where is the amplitude of the electric field ( ), is the permittivity of free space ( ), and is the speed of light ( ).
A simpler form using and , is where .
Also .
This can also be written using , so .
The standard formula is .
Given .
Distance from point source m.
For a point source radiating uniformly in all directions, the intensity at a distance is related to the power of the source by: Equating the two expressions for intensity: We know that .
So, .
Substitute the values: m, m/s, N/C.
This matches option (4).

Step 1: Use the de Broglie wavelength formula for an electron Step 2: Substitute the given potential

The general form of the electric field of an EM wave traveling in the -direction is: Given: The wave number Thus,

The intensity of light is given by: $ P r$ is the distance from the source. Substituting the given values, the intensity at a distance of 0.5 m is approximately 3.18 W/m².

The electric field is given as , which is in the form .
Angular frequency rad/s, wave number m .
Frequency Hz.
Speed of the wave m s .

The intensity of an electromagnetic wave is related to the rms value of the electric field by: where is the amplitude of the electric field ( ), is the permittivity of free space ( ), and is the speed of light ( ).
A simpler form using and , is where .
Also .
This can also be written using , so .
The standard formula is .
Given .
Distance from point source m.
For a point source radiating uniformly in all directions, the intensity at a distance is related to the power of the source by: Equating the two expressions for intensity: We know that .
So, .
Substitute the values: m, m/s, N/C.
This matches option (4).

Step 1: Electromagnetic Wave Relationship In an electromagnetic wave: where: - = electric field magnitude, - = magnetic field magnitude, - = speed of light ( m/s). Step 2: Given Condition The problem states that the magnetic field is scaled by : Since , we substitute: Dividing: Conclusion Thus, the correct answer is:

Step 1: Electromagnetic Wave Relationship In an electromagnetic wave: where: - = electric field magnitude, - = magnetic field magnitude, - = speed of light ( m/s). Step 2: Given Condition The problem states that the magnetic field is scaled by : Since , we substitute: Dividing: Conclusion Thus, the correct answer is:

Electromagnetic induction refers to the generation of an electromotive force (emf) in a conductor due to a changing magnetic field.

- Electric generator works on the principle of electromagnetic induction by converting mechanical energy into electrical energy via relative motion between a magnet and a coil.

- Galvanometer detects current but is not primarily based on electromagnetic induction; it operates using the magnetic effect of current.

- Electric motor converts electrical energy into mechanical energy using magnetic forces, not induction for emf generation.

- Ohm meter measures resistance using the flow of current, unrelated to electromagnetic induction.

Therefore, the device based on electromagnetic induction is the electric generator.

Power radiated by a linear antenna is inversely proportional to the square of the wavelength, i.e., Given the original power is at wavelength . At wavelength , the power radiated is: Thus, the power radiated at is .