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JEE Main Physics Class 12 Electromagnetic Waves Part 1 Questions
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© examsnet.com
Question : 41
Total: 100
The peak electric field produced by the radiation coming from the
8
W
bulb at a distance of
10
m
is
x
10
√
µ
0
c
π
V
m
. The efficiency of the bulb is
10
%
and it is a point source. The value of
x
is ............ .
[25 Feb 2021 Shift 2]
Your Answer:
Validate
Solution:
Given, radiation power
(
P
)
=
8
W
Distance
(
d
)
=
10
m
∵
Intensity
(
I
)
=
1
2
⊂
ε
0
E
2
=
Power
(
P
)
Area
(
A
)
. . . (i)
where,
c
=
1
√
µ
0
ε
0
=
speed of light in vacuum ... (ii)
and
E
=
electric field
From Eq. (ii), we get
ε
0
=
1
µ
0
c
2
Put this value of
ε
0
in Eq. (i), we get
I
=
1
2
C
1
µ
0
C
2
E
2
=
P
A
⇒
1
2
1
µ
0
C
E
2
=
P
A
⇒
E
=
√
2
P
µ
0
C
A
=
√
2
P
µ
0
C
4
π
d
2
=
√
2
P
4
d
2
√
µ
0
C
π
=
√
2
×
8
4
×
100
=
√
µ
0
C
π
=
2
10
√
µ
0
C
π
V
∕
m
∴
x
=
2
© examsnet.com
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