Electromagnetic Wave Amplitude Relationship Explained
This question asks us to find the relationship between the amplitude of the electric field ($E_0$) and the amplitude of the magnetic field ($B_0$) for a progressive electromagnetic wave. The electric field component is given by the formula $E = E_0\sin(kx - \omega t)$.
Understanding the Link Between Electric and Magnetic Field Amplitudes
For any electromagnetic wave travelling through a vacuum, there's a direct relationship between the amplitude of the electric field ($E_0$) and the amplitude of the magnetic field ($B_0$). This relationship involves the speed of light in a vacuum, represented by '$c$'. The electric field amplitude is precisely '$c$' times the magnetic field amplitude:
$ E_0 = c B_0 $
From this, we can express the ratio of the electric field amplitude to the magnetic field amplitude as:
$ \frac{E_0}{B_0} = c $
Connecting Wave Speed to Wave Parameters
The equation describing the electric field, $E = E_0\sin(kx - \omega t)$, provides details about the wave's characteristics:
- '$k$' is the wave number, indicating how many radians of phase change occur per unit distance.
- '$\omega$' is the angular frequency, indicating how fast the phase changes per unit time.
The speed of any wave ('$c$' in this case) is determined by its angular frequency ('$\omega$') and its wave number ('$k$'). The relationship is:
$ c = \frac{\omega}{k} $
Deriving the Amplitude Ratio Expression
By combining the two key relationships we've identified, we can find the answer. We know that $\frac{E_0}{B_0} = c$, and we also know that $c = \frac{\omega}{k}$. Substituting the second equation into the first gives us:
$ \frac{E_0}{B_0} = \frac{\omega}{k} $
This result tells us that the ratio of the electric field amplitude to the magnetic field amplitude is equal to the ratio of the wave's angular frequency to its wave number.
Comparing with the Given Options
Now, let's check which of the provided options matches our derived relationship:
- Option 1: $ \frac{E_0}{B_0} = \frac{\omega}{k} $
- Option 2: $ \frac{E_0}{B_0} = \frac{\omega^2}{k^2} $
- Option 3: $ \frac{E_0}{B_0} = \frac{k}{\omega} $
- Option 4: $ \frac{E_0}{B_0} = \frac{k^2}{\omega^2} $
Our calculated relationship, $ \frac{E_0}{B_0} = \frac{\omega}{k} $, exactly matches Option 1.
Important Questions from Electromagnetic Waves
Consider the two statements given below :
Statement-1: Infrared waves are also called heat waves.
Statement-2: Water molecules readily absorb infrared waves.
Select the correct answer using the code given below:
Consider the following statements about visible light, UV light and X-rays:
1. The wavelength of visible light is more than that of X-rays.
2. The energy of X-ray photons is higher than that of UV light photons.
3. The energy of UV light photons is less than that of visible light photons.
Which of the statements given above is/are correct?The wavelength of X-rays is of the order of
Which of the followings are the characteristics of electromagnetic waves?
1) They are elastic waves.
2) They can also move in a vacuum.
3) They have electric and magnetic components that are mutually perpendicular.
4) They move with a speed equal to 3 lakh meters per second.
Select the correct answer using the code given below:
Which of the following devices is based on the phenomenon of electromagnetic induction?
