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A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10-2 Am-1 at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given: Permeability of free space  speed of light in vacuum c = 3 × 108 ms-1)
  • a)
    16.96 Vm-1
  • b)
    2.25 × 10-2 Vm-1
  • c)
    8.48 Vm-1
  • d)
    6.75 × 106 Vm-1
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A plane electromagnetic wave travels in a medium of relative permeabil...
H = 4.5 × 10-2
So, B = μ0μH
Thus, E =  c/nB (where n ⇒ refractive index)
So, 

E = 8.48
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Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell’slaw,it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation,n= c /where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.In normal materials, both εr and μr , are positive, implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light.However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. Q.Choose the correct statement.

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell’slaw,it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation,n= c /where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity and permeability of the medium respectively.In normal materials, both εr and μr , are positive, implying positive n for the medium. When both εr and μr are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light.However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. Q.For light incident from air on a meta-material, the appropriate ray diagram is

Paragraph for Questions 13 and 14Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurringmaterial, then by Snell’s law, it is understood that the refracted ray bends towards the normal.But it never emerges on the same side of the normal as the incident ray. According to electromagnetism,the refractive index of the medium is given by the relation, where c is the speed ofelectromagnetic waves in vacuum, v its speed in the medium, εr and μr are the relative permittivity andpermeability of the medium respectively.In normal materials, both εr and μr, are positive, implying positive n for the medium. When both εr and μrare negative, one must choose the negative root of n. Such negative refractive index materials can nowbe artificially prepared and are called meta-materials. They exhibit significantly different optical behavior,without violating any physical laws. Since n is negative, it results in a change in the direction ofpropagation of the refracted light. However, similar to normal materials, the frequency of light remainsunchanged upon refraction even in meta-materials.Q.For light incident from air on a meta-material, the appropriate ray diagram is

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A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10-2Am-1at a point, what will be the approximate magnitude of electric field intensity at that point?(Given: Permeability of free spacespeed of light in vacuum c = 3 × 108ms-1)a)16.96 Vm-1b)2.25 × 10-2Vm-1c)8.48 Vm-1d)6.75 × 106Vm-1Correct answer is option 'C'. Can you explain this answer?
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A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10-2Am-1at a point, what will be the approximate magnitude of electric field intensity at that point?(Given: Permeability of free spacespeed of light in vacuum c = 3 × 108ms-1)a)16.96 Vm-1b)2.25 × 10-2Vm-1c)8.48 Vm-1d)6.75 × 106Vm-1Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10-2Am-1at a point, what will be the approximate magnitude of electric field intensity at that point?(Given: Permeability of free spacespeed of light in vacuum c = 3 × 108ms-1)a)16.96 Vm-1b)2.25 × 10-2Vm-1c)8.48 Vm-1d)6.75 × 106Vm-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10-2Am-1at a point, what will be the approximate magnitude of electric field intensity at that point?(Given: Permeability of free spacespeed of light in vacuum c = 3 × 108ms-1)a)16.96 Vm-1b)2.25 × 10-2Vm-1c)8.48 Vm-1d)6.75 × 106Vm-1Correct answer is option 'C'. Can you explain this answer?.
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