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JEE Main 24 June 2022 Shift 1 Solved Paper
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© examsnet.com
Question : 43
Total: 90
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
µ
0
=
4
π
×
10
−
7
NA
−
2
, speed of light in vacuum
c
=
3
×
10
8
ms
−
1
)
[24-Jun-2022-Shift-1]
16.96
Vm
−
1
2.25
×
10
−
2
Vm
−
1
8.48
Vm
−
1
6.75
×
10
6
Vm
−
1
Validate
Solution:
H
=
4.5
×
10
−
2
So
B
=
µ
0
µ
H
Thus
E
=
c
n
B
(where
n
⇒
refractive index)
So
E
=
3
×
10
8
×
4
π
×
10
−
7
×
1.61
×
4.5
×
10
−
2
√
1.61
×
6.44
E
=
8.48
© examsnet.com
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