Electromagnetic Waves MCQ Quiz - Objective Question with Answer for Electromagnetic Waves - Download Free PDF

CONCEPT:

The magnetic field at a distance 'r' from an infinitely long wire 

\(B =\frac{\mu_0}{4 \pi}\frac{2I}{r}\) (Where I is the current)

As shown in the figure the direction of B on the right side of the wire is inwards (⊗) and The direction of B on the left side of the wire is outwards (⊙)

EXPLANATION:

Let P be the point where the magnetic field becomes zero (as shown).

The distance from wires A, B, and C are (d-x), x, and (d+x)

So, the magnetic field due to A, B, and C

\(\begin{aligned} & ⇒B_A = \frac{\mu_0}{4 \pi} \frac{2I}{(d-x)} ⊗ \\ & ⇒B_B = \frac{\mu_0}{4 \pi} \frac{2I}{x} ⊙ \\ & ⇒B_C = \frac{\mu_0}{4 \pi} \frac{2I}{(d+x)}⊙ \\ \end{aligned}\)

∴ The total magnetic field at P 

\(\begin{aligned} &⇒ B_P = B_A + B_B+B_C \\ &⇒ B_P = \frac{\mu_0}{4 \pi} \frac{2I}{(d-x)} ⊗ + \frac{\mu_0}{4 \pi} \frac{2I}{x} ⊙ + \frac{\mu_0}{4 \pi} \frac{2I}{(d+x)}⊙ \\ & ⇒ B_P = \frac{2I\mu_0}{4 \pi} \left[ -\frac{1}{d-x} + \frac{1}{x}+ \frac{1}{d+x} \right] \\ & ⇒ B_P =\frac{2I\mu_0}{4 \pi} \left[ \frac{d^2 - 3x^2}{x(d-x)(d+x)} \right] \\ \end{aligned} \)

According to the problem, 

⇒ B_P = 0 (magnetic field vanishes)

\(\begin{aligned} & ⇒ \frac{2I\mu_0}{4 \pi} \left[ \frac{d^2 - 3x^2}{x(d-x)(d+x)} \right] = 0 \\ &∴ d^2 - 3x^2 = 0 \\ & x = \frac{d}{\sqrt{3}} \\ \end{aligned} \)

Like P there can be a similar point at \( x = -\frac{d}{\sqrt{3}}\) where also the magnetic field vanishes. 

So, the magnetic field vanishes at \( x = \pm \frac{d}{\sqrt{3}}\)

Hence the correct answer is option 4.

Calculation:
The pressure exerted by an electromagnetic wave on a surface can be calculated by the formula:

Pressure (P) = Energy / (Area × Time)

We are given:

• Energy = E
• Area = A
• Time = t
• Speed of light = c

Since the wave is perfectly reflected, the pressure is twice that of a non-reflected wave.

Therefore, the formula for the average pressure exerted on the surface is:

P = 2E / (A × c)

The correct answer is: 2E / (A × c)

Concept:

Behavior of Light at a Refractive Surface: When light passes from one medium to another, its speed and wavelength change, but its frequency remains constant.

• Relation of Speed, Wavelength, and Frequency:
• Formula: c = fλ
  • c = Speed of light
  • f = Frequency of light
  • λ = Wavelength of light
• Refractive Index and Wavelength Change:
• n = c_air / c_glass = λ_air / λ_glass
  • n = Refractive index of glass with respect to air
  • c_air = Speed of light in air
  • c_glass = Speed of light in glass

Calculation:

Given,

Wavelength in air: λ_air = 480 nm

Refractive index of glass: n = 1.5

⇒ The wavelength in glass:

λ_glass = λ_air / n

⇒ λ_glass = 480 nm / 1.5

⇒ λ_glass = 320 nm

⇒ Since frequency remains constant across media:

f_incident / f_refracted = 1 / 1

∴ The ratio of the frequency of incident and refracted light is 1:1.

Explanation:

The average value of the electric energy density in an electromagnetic wave is given by:

U = (1/4) ε₀ E₀²

This formula represents the average value of the electric energy density, where:

ε₀ is the permittivity of free space.

E₀ is the peak electric field strength in the electromagnetic wave.

The correct answer is option 4: (1/4) ε₀ E₀².

Concept:

• Electric and Magnetic Energy Density:
• The total energy density of an electromagnetic wave is the sum of the energy densities of the electric and magnetic fields.
  • U = U_E + U_B, where:
    • U_E: Electric energy density
    • U_B: Magnetic energy density
• For an electromagnetic wave, the electric and magnetic energy densities are equal:
  • U_E = U_B = (1/2)ε₀E₀² = (B₀²) / (2μ₀)
• The ratio of average electric energy density to total energy density is:
  • U_E / U = 1/2

Calculation:

We have:

U_E / U = U_E / (U_E + U_B) = U_E / (2U_E) = 1/2

∴ The ratio of average electric energy density to total energy density is 1/2.

The correct answer is Microwaves.Key Points

• The option with the highest wavelength among the given electromagnetic waves is microwaves.
• Microwaves have wavelengths ranging from 1 millimeter to 1 meter.
• Microwaves move in a straight line.
• Within the electromagnetic spectrum, microwaves have a frequency that is higher than that of regular radio waves and lower than that of infrared light.
• They are employed in radar, communications, radio astronomy, remote sensing, and, of course, cooking because of their heating function.

Additional Information

• Infrared waves have wavelengths ranging from 700 nanometers to 1 millimeter.
  • They are commonly used in heating and communication applications.
• Ultraviolet waves have wavelengths ranging from 10 nanometers to 400 nanometers.
  • They are commonly used in sterilization and detection applications.
• X-rays have wavelengths ranging from 0.01 nanometers to 10 nanometers.
  • They are commonly used in medical imaging and material analysis applications.

Calculation:
The pressure exerted by an electromagnetic wave on a surface can be calculated by the formula:

Pressure (P) = Energy / (Area × Time)

We are given:

• Energy = E
• Area = A
• Time = t
• Speed of light = c

Since the wave is perfectly reflected, the pressure is twice that of a non-reflected wave.

Therefore, the formula for the average pressure exerted on the surface is:

P = 2E / (A × c)

The correct answer is: 2E / (A × c)

Concept:

In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of propagation. The relationship between the amplitudes of the electric and magnetic fields in a plane electromagnetic wave is given by:

c = E/B

where:

• E is the electric field amplitude,
• B is the magnetic field amplitude, and
• c is the speed of light (approximately 3 × 10⁸ m/s).

From the given magnetic field expression:

B = 2 × 10^-8 sin(0.5 × 10³ x + 1.5 × 10¹¹ t) T

We can identify the amplitude of the magnetic field, B₀ = 2 × 10^-8 T.

Now, using the formula c = E/B, we can find the electric field amplitude:

E = c × B

E = (3 × 10⁸ m/s) × (2 × 10^-8 T)

E = 6 V/m

The direction of the electric field is along the x-axis because the magnetic field is along the y-axis, and the wave is traveling along the z-axis (perpendicular to both). Thus, the electric field is along the x-axis.

∴ The correct answer is option 1: 6 V/m along x-axis.

Concept:

Behavior of Light at a Refractive Surface: When light passes from one medium to another, its speed and wavelength change, but its frequency remains constant.

• Relation of Speed, Wavelength, and Frequency:
• Formula: c = fλ
  • c = Speed of light
  • f = Frequency of light
  • λ = Wavelength of light
• Refractive Index and Wavelength Change:
• n = c_air / c_glass = λ_air / λ_glass
  • n = Refractive index of glass with respect to air
  • c_air = Speed of light in air
  • c_glass = Speed of light in glass

Calculation:

Given,

Wavelength in air: λ_air = 480 nm

Refractive index of glass: n = 1.5

⇒ The wavelength in glass:

λ_glass = λ_air / n

⇒ λ_glass = 480 nm / 1.5

⇒ λ_glass = 320 nm

⇒ Since frequency remains constant across media:

f_incident / f_refracted = 1 / 1

∴ The ratio of the frequency of incident and refracted light is 1:1.

Concept:

The relationship between the electric field E_o The relationship between the electric field B_o in an electromagnetic wave propagating through a vacuum is governed by the equation derived from Maxwell's equations.

Calculation: 

For an electromagnetic wave in a vacuum, the speed of the wave c  is related to E_o and B_o as

\(c = \frac{E_o}{B_o}\)

Also c = \(\frac{ω}{k}\)

⇒ \(\frac{ω}{k} = \frac{E_o}{B_o}\)

⇒ kE_o = ωB_o

∴ the correct relation between E0 and B0 is kEo = ωBo

The correct answer is Microwaves.Key Points

• The option with the highest wavelength among the given electromagnetic waves is microwaves.
• Microwaves have wavelengths ranging from 1 millimeter to 1 meter.
• Microwaves move in a straight line.
• Within the electromagnetic spectrum, microwaves have a frequency that is higher than that of regular radio waves and lower than that of infrared light.
• They are employed in radar, communications, radio astronomy, remote sensing, and, of course, cooking because of their heating function.

Additional Information

• Infrared waves have wavelengths ranging from 700 nanometers to 1 millimeter.
  • They are commonly used in heating and communication applications.
• Ultraviolet waves have wavelengths ranging from 10 nanometers to 400 nanometers.
  • They are commonly used in sterilization and detection applications.
• X-rays have wavelengths ranging from 0.01 nanometers to 10 nanometers.
  • They are commonly used in medical imaging and material analysis applications.

Concept:

• The transmission coefficient (T) for an electromagnetic wave moving from a vacuum into a medium with refractive index (n) is given by: \(T = \frac {2n_1} {(n_1+n_2)}\)
• This coefficient (T) is also expressed as the ratio of the amplitude of the transmitted wave (E_t) to the amplitude of the incident wave (E₀): \(T = \frac {E_t}{E_0}\)

Explanation:

• The refractive index of the vacuum is : \(n_1=1\)
• In the problem, it states that \(E_t = \frac {2E_0}{3}\).
• Substituting the value in the formula of the coefficient of transmission: T

\(\therefore \frac 23=\frac{2n_1}{n_1+n_2}\)

\(\therefore\frac 23=\frac{2}{1+n_2}\)

\(\therefore\frac 13=\frac{1}{1+n_2}\)

\(\therefore 1+n_2=3\)

\(\therefore n_2=2\)

Given:

The Solar Radiation is,

1) Stationary wave

2) Mechanical wave

3) Transverse EM wave

4) Longitudinal EM wave

The correct answer is option 3.

Concept:

Solar radiation consists of electromagnetic waves. Electromagnetic waves are waves that are created as a result of vibrations between an electric field and a magnetic field. They are transverse in nature, meaning the oscillations are perpendicular to the direction of the wave's advance.

Calculation:

We know that:

⇒ Solar radiation is electromagnetic radiation

⇒ Electromagnetic radiation consists of transverse waves

⇒ Therefore, solar radiation is a transverse electromagnetic wave.

∴ Solar radiation is a transverse EM wave.

Concept:

The concept used in this question is the relationship between energy, momentum, and the speed of light in the case of electromagnetic waves.

When electromagnetic radiation (such as light) is absorbed by a surface, the momentum delivered to the surface can be calculated using the relationship between energy and momentum for light.

The formula to relate energy E and momentum P for light is

\(p = \frac{E}{c}\)

Calculation: 

Given

E =  6.48 × 105 

c = 3 × 10⁸

\(\mathrm{p}=\frac{\mathrm{E}}{\mathrm{C}}=\frac{6.48 × 10^5}{3 × 10^8}=2.16 × 10^{-3}\)

∴ the total momentum delivered to the surface is 2.16 × 10-3 kg m/s

Explanation:

The volume charge density ρ, giving rise to a electrostatic potential Φ, can be found using Poisson's equation in electrostatics, which states that: \(∇²Φ = \frac{-ρ}{ε_₀}\), where ∇² is the Laplacian operator representing the divergence of the gradient (a measure of the rate at which the potential changes).

In Cartesian coordinates, the Laplacian operator ∇² takes the form: \(∇²Φ = \frac{∂²Φ}{∂x²} + \frac{∂²Φ}{∂y²} + \frac{∂²Φ}{∂z²}.\)

Given \( Φ = k(x² + y² + z²),\) we find ∇²Φ by taking the second derivatives with respect to x, y, and z and adding the results:

\(∇²Φ = \frac{∂²Φ}{∂x²} + \frac{∂²Φ}{∂y²} + \frac{∂²Φ}{∂z²} = 2k + 2k + 2k = 6k.\)

So, substituting ∇²Φ = 6k into Poisson's equation gives: 6k = -ρ/ε₀, Implying that ρ = -6kε₀. So, the volume charge density ρ giving rise to the potential Φ is -6kε₀.