Animated Solution for Physics - Electromagnetic Waves: The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by
E=30(2x^+y^)sin[2π(5×1014t−3107z)] V m−1. Which of the following option(s) is(are) correct?
[Given: The speed of light in vacuum, c=3×108 ms−1]
Select Answer:
* Multiple Correct
Visualized Solution
\text{Visualizing the EM Wave}
E=30(2i^+j^)sin[2π(5×1014t−3107z)]
Propagation direction: c^=k^
\text{Extracting Wave Parameters}
Standard form: E=E0sin(ωt−kz)
ω=2π×5×1014=1015π rad/s
k=2π×3107=32π×107 m−1
\text{Wave Speed in Medium}
v=kω
v=32π×1071015π=1.5×108 m/s
\text{Refractive Index}
n=vc
n=1.5×1083×108=2
⇒Option (D) is correct
\text{Polarization Angle}
E0=60i^+30j^
tanθ=ExEy=6030=21
θ=tan−1(21)=30∘
⇒Option (C) is incorrect
\text{Magnetic Field Amplitude}
B0=vE0
E0=∣E0∣=3022+12=305 V/m
B0=1.5×108305=25×10−7 T
\text{Magnetic Field Direction}
B^=c^×E^
B^=k^×52i^+j^
B^=52(k^×i^)+(k^×j^)=52j^−i^
\text{Final Magnetic Field Vector}
B=B0B^sin(ωt−kz)
B=25×10−7(5−i^+2j^)sin(…)
B=2×10−7(−i^+2j^)sin(…)
Bx=−2×10−7sin(…)
\text{The Way Forward}
Energy Density: u=21ϵE02=2μB02
Poynting Vector: S=μ1(E×B)
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The Sigma Insight: Characteristics of Electromagnetic Waves
Solution Diagram
Decoding the Wave Equation
When you first look at the electric field equation of an electromagnetic wave, it might seem like a dense jungle of numbers and variables. But fear not! The key is to compare it with the standard wave equation:
E=E0sin(ωt−kz)
In our problem, the phase is given as 2π(5×1014t−3107z). To properly extract the angular frequency ω and the wave number k, we must distribute the 2π inside the bracket.
This gives us ω=1015π rad/s and k=32π×107 m−1. Also, the negative sign between the time and space terms tells us that the wave is propagating in the positive z-direction. So, our propagation unit vector is c^=k^.
The Speed Limit of the Medium
Now that we have ω and k, we hold the keys to finding the speed of the wave in this specific dielectric medium. The wave speed v is simply the ratio of the angular frequency to the wave number:
v=kω=32π×1071015π=1.5×108 m/s
With the speed in the medium known, calculating the refractive index n is a breeze. It is the ratio of the speed of light in vacuum c to the speed in the medium v:
n=vc=1.5×1083×108=2
This perfectly matches Option (D)!
The Geometry of Polarization
What about the polarization of the wave? An electromagnetic wave is polarized along the direction of its electric field vector. Here, the electric field vector is E0=30(2i^+j^)=60i^+30j^.
To find the angle θ this vector makes with the x-axis, we take the tangent, which is the ratio of the y-component to the x-component:
tanθ=ExEy=6030=21
Since tan(30∘)=1/3, our angle θ=tan−1(0.5) is definitely not 30∘ (it's actually about 26.56∘). Therefore, Option (C) is incorrect.
Unveiling the Magnetic Field
To find the magnetic field, we need two things: its amplitude and its direction. The amplitude B0 is directly related to the electric field amplitude E0 and the wave speed v:
B0=vE0
First, let's find the magnitude of the electric field: E0=602+302=305 V/m.
Plugging this into our relation gives:
B0=1.5×108305=25×10−7 T
The Cross Product Magic
Finally, we must determine the direction of the magnetic field. In an electromagnetic wave, the electric field, magnetic field, and propagation direction form a mutually orthogonal right-handed system. The direction of the magnetic field B^ is given by the cross product:
B^=c^×E^
Substituting our known unit vectors:
B^=k^×52i^+j^=52(k^×i^)+(k^×j^)=52j^−i^
Now, we assemble the full magnetic field vector by multiplying the amplitude, the direction vector, and the sine phase factor:
B=25×10−7(5−i^+2j^)sin(…)
Notice how beautifully the 5 cancels out! We are left with:
B=2×10−7(−i^+2j^)sin(…)
From this, we can clearly see that the x-component is Bx=−2×10−7sin(…), which makes Option (A) correct. However, the y-component is By=4×10−7sin(…), which means Option (B) is incorrect.
