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JEE Main 9 Jan 2020 Shift 1 Solved Paper
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© examsnet.com
Question : 4
Total: 75
The electric fields of two plane electromagnetic plane waves in vacuum are given by
→
E
1
=
E
0
^
j
cos
(
ω
t
−
k
x
)
and
→
E
2
=
E
0
^
k
cos
(
ω
t
−
k
y
)
At t = 0, a particle of charge q is at origin with a velocity
→
v
=
0.8
c
^
j
(c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
E
0
q
(
−
0.8
^
i
+
^
j
+
^
k
)
E
0
q
(
0.8
^
i
−
^
j
+
0.4
^
k
)
E
0
q
(
0.8
^
i
+
^
j
+
0.2
^
k
)
E
0
q
(
0.4
^
i
−
3
^
j
+
0.8
^
k
)
Validate
Solution:
→
E
1
=
E
0
^
j
cos
(
ω
t
−
k
x
)
Its corresponding magnetic field will be
→
B
1
=
E
0
c
^
k
cos
(
ω
t
−
k
x
)
→
E
2
=
E
0
^
k
cos
(
ω
t
−
k
y
)
→
B
2
=
E
0
c
^
i
cos
(
ω
t
−
k
y
)
Net force on charge particle
=
q
→
E
1
+
q
→
E
2
+
q
→
v
×
→
B
1
+
q
→
v
×
→
B
2
=
q
E
0
^
j
+
q
E
0
^
k
+
q
(
0.8
c
^
j
)
×
(
E
0
c
^
k
)
+
q
(
0.8
c
^
j
)
×
(
E
0
c
^
i
)
→
F
=
E
0
q
(
0.8
^
i
+
^
j
+
0.2
^
k
)
© examsnet.com
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