Electromagnetic Waves

Energy is equally divided between the electric and magnetic fields. The Electric energy density is equivalent to the magnetic energy density.

Here, u_E= ½ ϵ₀E²

u_B=B²/2μ0

Also, E=cB and c=1/√μ₀ϵ₀

Therefore, we get u_E = ½ ϵ₀E² = ½ ϵ₀(cB)²

=½ ϵ₀ B²

u_E= u_B

So, the correct answer is C) During the propagation of electromagnetic waves in a medium, electric energy density is equal to the magnetic energy density.

To solve this problem, we need to determine the relationship between the relative permittivities of air ( ) and the medium ( ) given the electromagnetic (EM) wave characteristics.

An electromagnetic wave undergoes a change in speed when it enters a different medium. The speed of the EM wave in a medium is given by:

where is the speed of light in a vacuum, and is the relative permittivity of the medium.

From the problem, we are given the electric field equations:

• In air: , with wave number in air as .
• In the medium: , with wave number as .

By comparing the wave numbers in both media, we have:

• In air: .
• In the medium: (as speed decreases).

Given that the wave number is maintained, , we have:

Thus, by using the formula for the change in wavelength, we get:

Substitute the given condition of into the formula:

This indicates:

Hence, the correct option is:

To solve the given problem, let's start by reviewing the fundamental concepts of electromagnetic (EM) waves and their properties:

• Electromagnetic waves are characterized by oscillating electric and magnetic fields, which are perpendicular to each other and to the direction of wave propagation.
• The general form of an electromagnetic wave propagating in the x-direction is given by the equations: for the electric field, and for the magnetic field.
• In a vacuum, the speed of electromagnetic waves, , is approximately , which relates the wave's frequency and wavelength.

Given data in the problem:

• Wavelength
• Maximum electric field magnitude
• The wave is propagating in a vacuum along the x-direction.

First, let's find the wave number and the angular frequency :

• Wave number, : It is given by .
  • Substitute the given wavelength:
• Angular frequency, : As the wave is in a vacuum, .
  • Substitute the speed of light:

Given options, the correct equations for such a wave are chosen under the following considerations:

• Amplitude of electric field
• The relationship between electric and magnetic fields in a vacuum is .

Calculate :

Therefore, the correct option is:

The correct answer is (B) : λ_gamma-ray < λ_X-ray < λ_visible < λ_microwave

Explanation:

Wave length of microwave is maximum then visible light then X-rays and then gamma rays.
so the correct order will be

The correct answer is (B) : (B) & (D) only
In an EM wave :
1.
2.
3. Energy is equally divided
4.

The correct solution is (B) : (A) & (E) only


varies linearly with frequency.
And, threshold wavelength can be explained by particle nature of light.

The correct answer is (A) : 0.2
Modulation Index ,

H = 4.5 × 10^–2
So B = μ_0μH
Thus
E= B
(where n ⇒ refractive index)
So
E =
E = 8.48
The correct option is (A): 16.96 Vm^–1

The correct answer is (B) : 4.0 × 10^–5 s
E₀ = 2.25 V/m
B₀ = 1.5 × 10^–8T
m/s
⇒ Refractive index = 2
Distance to be travelled = 6 km
Time taken
s

= 3.5×10⁹ Hz

∴ λ=

λ= =

∴ Size of antenna =

=

=

The correct option is (C) : 21.4 mm

To solve the question of matching List-I with List-II, we need to understand the application of each type of wave or ray listed in List-I:

1. UV rays: These are commonly used in water purification systems because they have the ability to kill bacteria and viruses present in the water. UV rays can penetrate microorganisms and destroy their DNA, making them an effective method for disinfecting water. Thus, the correct match is with (ii) Water purification.
2. X-rays: These rays are widely used as a diagnostic tool in medicine. X-rays can penetrate tissues and create images of the inside of the body, which are used for diagnosing fractures, infections, and other medical conditions. Therefore, X-rays best match with (i) Diagnostic tool in medicine.
3. Microwave: Microwaves are used in communication and radar systems. They can transmit signals over long distances and are crucial for modern communication systems, including satellite transmission. Hence, microwaves correctly pair with (iii) Communication, Radar.
4. Infrared wave: These waves are used for improving visibility in foggy conditions. Infrared waves can penetrate fog effectively, which enhances visibility. As such, Infrared waves correspond to (iv) Improving visibility in foggy days.

Based on the reasoning above, the correct matching of the lists is:

Therefore, the correct answer to the given question is: (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv).

The correct answer is (D): 10

Explanation:

Modulate signal s(t) ≡ [1 + 20sin(1000πt)] sin(10⁷πt)
≡ sin(10⁷πt) + 10cos(10⁷πt – 10³πt) + 10cos(10⁷πt + 10³πt)
⇒ Required amplitude = 10

To determine the total bandwidth of the given amplitude modulated wave, we begin by understanding the formula for an amplitude modulated (AM) signal. The given equation is:

where:

• The carrier frequency, , is Hz or 10 MHz.
• The modulating frequency, , is Hz or 10 kHz.

The general formula for an AM wave is:

where is the modulation index.

The total bandwidth of an AM signal is given by:

Given that the modulating frequency , the total bandwidth is calculated as:

Thus, the total bandwidth of the amplitude modulated wave is 20 kHz, which matches the given correct answer.

Ratio

Ratio

Therefore, The correct answer is (C) :

The correct answer is (C) : 375
Let's consider H' as the antenna's new height.
As the range of antenna ,
Hence , d₁ = d = ……… (i)
d₂ = 2d = ……….. (ii)
By dividing (i) and (ii) , we get

Then , squaring the above equation , we get

H' = 500 m
∴ Increase in height = H' - H
= 500 - 125
= 375 m

To determine the extension of the wire when a transverse wave passes through, we first find the tension (T) in the wire due to the wave's speed (v).

The wave speed formula is: v = √(T/μ), where μ is the linear mass density (mass per unit length).
Linear mass density, μ = (mass of string)/(length of string) = (10 g)/(50 cm) = (10 × 10⁻³ kg)/(0.50 m) = 0.02 kg/m.
Substitute v and μ into the wave speed formula:

60 m/s = √(T/0.02 kg/m).
Squaring both sides:
(60)² = T/0.02.
T = (60)² × 0.02 = 72 N.
The tension in the wire is 72 N.

Next, we determine the extension (ΔL) using Young's modulus (Y). The formula is: ΔL = (FL)/(AY), where F is the force (tension), L is the original length, A is the cross-sectional area, and Y is Young's modulus.
Substituting values:
ΔL = (72 N × 0.50 m)/(2 × 10⁻⁶ m² × 1.2 × 10¹¹ N/m²).
ΔL = 72 × 0.50/(2.4 × 10⁵)
ΔL = 36/240000.
ΔL = 1.5 × 10⁻⁴ m.

Hence, x is given as:
ΔL = x × 10⁻⁵ m
1.5 × 10⁻⁴ m = x × 10⁻⁵ m.
x = 15.

The value of x is 15, which falls within the expected range of (15, 15).