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In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field is
- a)903 sin (0.83x) sin (108 t) uy V/m
- b)903 sin (1.2x) sin (108 t) uy V/m
- c)903 sin (0.83x) cos (108 t) uy V/m
- d)903 sin (1.2x) cos (108 t) uy V/m
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wav...
Given:
- Magnetic field of an EM wave in a non-magnetic medium is given as H = 6 cos β × cos (108 t) us A/m
- ∊r (relative permittivity) = 6.25
To find:
The corresponding electric field
Solution:
The relationship between the magnetic field (H) and the electric field (E) in an electromagnetic wave is given by:
E = (1/√(∊r)) * H
Substituting the given values:
E = (1/√(6.25)) * 6 cos β × cos (108 t) us A/m
Simplifying the expression:
E = (6/√(6.25)) cos β × cos (108 t) us A/m
E = (6/2.5) cos β × cos (108 t) us A/m
E = 2.4 cos β × cos (108 t) us A/m
The electric field is given as E = E0 sin (kx - ωt) uy V/m, where E0 is the amplitude of the electric field, k is the wave number, x is the position coordinate, ω is the angular frequency, and t is the time.
Comparing the given electric field expression with the standard form:
E0 sin (kx - ωt) uy V/m
We can see that:
- Amplitude E0 = 903 V/m (approximated value of 2.4)
- Wave number k = 0.83 (approximated value of β)
- Angular frequency ω = 108
Therefore, the corresponding electric field is:
E = 903 sin (0.83x) sin (108 t) uy V/m
Hence, the correct answer is option 'A': 903 sin (0.83x) sin (108 t) uy V/m.
- Magnetic field of an EM wave in a non-magnetic medium is given as H = 6 cos β × cos (108 t) us A/m
- ∊r (relative permittivity) = 6.25
To find:
The corresponding electric field
Solution:
The relationship between the magnetic field (H) and the electric field (E) in an electromagnetic wave is given by:
E = (1/√(∊r)) * H
Substituting the given values:
E = (1/√(6.25)) * 6 cos β × cos (108 t) us A/m
Simplifying the expression:
E = (6/√(6.25)) cos β × cos (108 t) us A/m
E = (6/2.5) cos β × cos (108 t) us A/m
E = 2.4 cos β × cos (108 t) us A/m
The electric field is given as E = E0 sin (kx - ωt) uy V/m, where E0 is the amplitude of the electric field, k is the wave number, x is the position coordinate, ω is the angular frequency, and t is the time.
Comparing the given electric field expression with the standard form:
E0 sin (kx - ωt) uy V/m
We can see that:
- Amplitude E0 = 903 V/m (approximated value of 2.4)
- Wave number k = 0.83 (approximated value of β)
- Angular frequency ω = 108
Therefore, the corresponding electric field is:
E = 903 sin (0.83x) sin (108 t) uy V/m
Hence, the correct answer is option 'A': 903 sin (0.83x) sin (108 t) uy V/m.
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Community Answer
In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wav...
= 903 sin (0.83x) sin (108t)uy V/m
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In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer?
Question Description
In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer?.
In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer?.
Solutions for In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE).
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ample number of questions to practice In a non-magnetic medium ( ∊r= 6.25), the magnetic field of an EM wave is H = 6 cos β × cos (108 t) us A/m. The corresponding electric field isa)903 sin (0.83x) sin (108 t) uy V/mb)903 sin (1.2x) sin (108 t) uy V/mc)903 sin (0.83x) cos (108 t) uy V/md)903 sin (1.2x) cos (108 t) uy V/mCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
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