The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by \(\vec{E}\)=30(2π₯Μ + π¦Μ)sin [2π (5Γ1014π‘ β\(\frac{10^7}{3}z\))] Vmβ1. Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, π = 3 Γ 108 msβ1]
- π΅π₯ = β2 Γ 10β7 sin [2π(5Γ1014π‘β\(\frac{10^7}{3}z\))]Wbβ2
- π΅y=2Γ10β7 sin[2π(5Γ1014π‘β\(\frac{10^7}{3}z\))]Wbβ2
- The wave is polarized in the π₯π¦-plane with a polarization angle 30Β° with respect to the π₯-axis.
- The refractive index of the medium is 2
The Correct Option is A, D
Approach Solution - 1
\(v=\frac{3\times5\times10^{14}}{10^7}\)
\(v=1.5\times10^8\)
refractive index = \(\mu\) = \(\frac{C}{V}=\frac{3\times10^8}{1.5\times10^8}=2\)
\(\mu=2\)
given, \(\vec{E}=30(2\hat{x}+\hat{y})\,sin(2\pi(5\times10^{14}-\frac{10^7}{3}))^\frac{1}{m}\)
\(B_0=\frac{E_0}{V}=\frac{30\sqrt{5}}{1.5\times10^8}\)
Direction of \(\vec{B_0}\) is (\(\vec{V}\times\vec{E}\))
\(\vec{V}\times\vec{E}=\hat{k}\times\frac{2\hat{i}+\hat{j}}{\sqrt5}\)
\((\frac{-\hat{i}+2\hat{j}}{\sqrt5})\) put value \(\vec{B_0}\) = \(\frac{30\sqrt5}{1.5\times10^8}\times(\frac{-\hat{i}+2\hat{j}}{\sqrt5})\)
\(B_x=-2\times10^7\)
Approach Solution -2
Explanation:
- The speed of light \( V \) in a medium is given by \( \frac{\omega}{k} \).
- Using the provided frequency and wave number, \( v = \frac{3 \times 5 \times 10^{14}}{10^7} \).
- This calculates to \( v = 1.5 \times 10^8 \) m/s.
- The refractive index \( \mu \) is calculated as \( \frac{C}{V} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2 \).
- Given the electric field \( \vec{E} = 30(2\hat{x}+\hat{y})\sin\left[2\pi\left(5\times10^{14}t-\frac{z}{3\times10^7}\right)\right] \, \text{Vm}^{-1} \).
- The magnetic field amplitude \( B_0 = \frac{E_0}{V} = \frac{30\sqrt{5}}{1.5\times10^8} \).
- The direction of \( \vec{B_0} \) is given by \( \vec{V} \times \vec{E} \).
- Calculating \( \vec{V} \times \vec{E} = \hat{k} \times \frac{2\hat{i} + \hat{j}}{\sqrt{5}} = \frac{-\hat{i} + 2\hat{j}}{\sqrt{5}} \).
- Therefore, \( \vec{B_0} = \frac{30\sqrt{5}}{1.5\times10^8} \times \frac{-\hat{i}+2\hat{j}}{\sqrt{5}} \).
- Finally, \( B_x = -2 \times 10^{-7} \).
Further:
- The speed of light \( V \) in the medium is derived using \( V = \frac{\omega}{k} \).
- Calculating with the given values, \( V = 1.5 \times 10^8 \, \text{m/s} \).
- The refractive index \( \mu \) is found to be 2 using \( \mu = \frac{C}{V} \).
- Given the electric field expression, the magnetic field \( B_0 \) and its direction are determined by \( \vec{V} \times \vec{E} \).
- The correct options are A and D.
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Electromagnetic waves
The waves that are produced when an electric field comes into contact with a magnetic field are known as Electromagnetic Waves or EM waves. The constitution of an oscillating magnetic field and electric fields gives rise to electromagnetic waves.
Types of Electromagnetic Waves:
Electromagnetic waves can be grouped according to the direction of disturbance in them and according to the range of their frequency. Recall that a wave transfers energy from one point to another point in space. That means there are two things going on: the disturbance that defines a wave, and the propagation of wave. In this context the waves are grouped into the following two categories:
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