Animated Solution for Physics - Electromagnetic Waves: The electric field associated with an electromagnetic wave travelling in vacuum is given by E=E0sin(3y+4z+ωt)i^, where ω is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are:
[Given: speed of light in vacuum c=3×108 ms−1.]
Select Answer:
* Multiple Correct
Visualized Solution
Standard Equation of EM Wave
E=E0sin(ωt−k⋅r)
Given: E=E0sin(3y+4z+ωt)i^
Rewrite phase: ωt−(−3y−4z)
Wave Vector k
k⋅r=−3y−4z
k=−3j^−4k^
k=(−3)2+(−4)2=5 m−1
Direction of Propagation
v^=kk=5−3j^−4k^
v^=−51(3j^+4k^)
Option (A) is Correct.
Angular Frequency ω
c=kω⟹ω=c⋅k
ω=(3×108)×5
ω=1.5×109 rad/s
Option (C) is Correct.
Magnetic Field Direction
B^=v^×E^
B^=[−51(3j^+4k^)]×i^
B^=−51[3(j^×i^)+4(k^×i^)]
B^=5−4j^+3k^
Magnetic Field Equation
B0=cE0
B=cE0sin(3y+4z+ωt)(5−4j^+3k^)
Option (D) is Incorrect.
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The Sigma Insight: Characteristics of Electromagnetic Waves
Solution Diagram
The Anatomy of an Electromagnetic Wave
Electromagnetic waves are a beautiful manifestation of oscillating electric and magnetic fields propagating through space. To truly understand a wave, we must decode its mathematical signature. The standard equation for the electric field of an EM wave traveling in a vacuum is given by:
E=E0sin(ωt−k⋅r)
Here, E0 is the amplitude vector, ω is the angular frequency, and k is the wave vector which points in the direction of propagation. The term (ωt−k⋅r) is the phase of the wave.
Decoding the Phase
In our problem, the electric field is given as:
E=E0sin(3y+4z+ωt)i^
At first glance, the phase (3y+4z+ωt) looks slightly different from our standard form. To extract the wave vector k accurately, we can rewrite the phase to match the standard template (ωt−k⋅r):
ωt−(−3y−4z)
By comparing this with ωt−(kxx+kyy+kzz), we can immediately identify the components of the wave vector:
k=−3j^−4k^
Finding the Wave Vector and Direction
The magnitude of the wave vector, often called the wave number k, is crucial. It is calculated as:
k=(−3)2+(−4)2=9+16=5 m−1
This immediately tells us that Option (B) is incorrect, as it claims the magnitude is 0.5 m−1.
The direction of wave propagation is simply the unit vector along k, denoted as v^:
v^=kk=5−3j^−4k^=−51(3j^+4k^)
This perfectly matches Option (A), confirming it as a correct statement.
The Speed of Light Connection
The wave number k and the angular frequency ω are intimately connected by the speed of the wave, which in a vacuum is the speed of light c:
c=kω
Rearranging this to solve for ω, we get:
ω=c⋅k=(3×108 m/s)×(5 m−1)=1.5×109 rad/s
This confirms that Option (C) is also correct.
The Magnetic Field Cross Product
In an electromagnetic wave, the electric field E, the magnetic field B, and the direction of propagation v^ form a mutually orthogonal, right-handed system. The direction of the magnetic field is given by the cross product:
B^=v^×E^
Substituting our known unit vectors:
B^=[−51(3j^+4k^)]×i^
Using the cyclic properties of cross products (j^×i^=−k^ and k^×i^=j^):
B^=−51[3(−k^)+4(j^)]=53k^−4j^=5−4j^+3k^
The amplitude of the magnetic field is B0=cE0. Combining the amplitude and the direction, the full magnetic field vector is:
B=cE0sin(3y+4z+ωt)(5−4j^+3k^)
Looking at Option (D), it suggests the direction is (4j^−3k^) and completely misses the 51 factor. Therefore, Option (D) is incorrect.
