Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
CBSE
CBSE Question Papers
NCERT Books
NCERT Exemplar Books
NCERT Study Notes
CBSE Study Concepts
CBSE Class 10 Solutions
CBSE Class 12 Solutions
NCERT Text Book Class 11 Solutions
NCERT Text Book Class 12 Solutions
ICSE Class 10 Papers
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Contact Us
Our Apps
Privacy
Test Index
AP EAPCET 07-Jul-2022 Shift 1 Paper
Show Para
Hide Para
Section:
Physics
Share question:
© examsnet.com
Question : 115
Total: 160
The equation that represents magnetic field of a plane electromagnetic wave which is propagating along
x
-direction with wavelength
10
m
m
and maximum electric field
60
V
m
−
1
in
y
-direction is (c - speed of light)
(
6
×
10
−
7
)
‌
sin
[
0.2
Ï€
(
c
t
−
x
)
]
‌
∧
k
tesla
(
2
×
10
−
7
)
‌
sin
[
200
Ï€
(
c
t
−
x
)
]
‌
∧
k
tesla
(
2
×
10
−
7
)
‌
sin
[
200
Ï€
(
c
t
−
x
)
]
‌
∧
i
tesla
(
6
×
10
−
7
)
‌
sin
[
0.2
Ï€
(
c
t
−
x
)
]
‌
∧
ı
tesla
Validate
Solution:
Equation of magnetic field of an electromagnetic wave is given by :
B
=
B
0
‌
sin
(
ω
t
−
k
x
)
‌
∧
k
=
B
0
‌
sin
(
2
Ï€
f
t
−
2
Ï€
λ
.
x
)
‌
∧
k
.
.
.
[
∵
ω
=
2
Ï€
f
‌
and
‌
k
=
2
Ï€
λ
]
B
0
‌
sin
‌
2
Ï€
λ
[
f
λ
t
−
x
]
‌
∧
k
B
=
B
0
‌
sin
‌
2
Ï€
λ
[
c
t
−
x
]
‌
∧
k
.
.
.
(
i
)
Also direction of
B
vector is perpendicular to
E
and also tp direction of propagation
And
B
0
=
E
0
c
Here,
E
0
=
60
Vm
−
1
∴
B
0
=
60
3
×
10
8
=
20
×
10
−
7
Also
k
=
2
Ï€
λ
Here
k
=
200
Ï€
=
2
Ï€
λ
⇒
λ
=
1
100
m
=
100
100
‌
cm
=
10
‌
mm
=
10
−
2
m
Putting the values is eq. (i), we get
B
=
2
×
10
=
−
7
‌
sin
[
2
Ï€
10
−
2
(
c
t
−
x
)
]
‌
∧
k
=
2
×
10
−
7
‌
sin
‌
200
‌
Ï€
(
t
−
x
)
‌
∧
k
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
Prev Question
Next Question
More Free Exams:
AP and TS EAMCET Engineering Model Papers
AP EAMCET Engineering Previous Papers
AP EAPCET Chapters wise Questions
BITSAT Exam Previous Papers
KVPY Exam Previous Papers
TG EAMCET Engineering Previous Papers
TG EAPCET Chapters wise Questions
VITEEE Previous Papers