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JEE Main 7 Jan 2020 Shift 2 Solved Paper
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© examsnet.com
Question : 5
Total: 75
The electric field of a plane electromagnetic wave is given by
→
E
=
E
0
^
i
+
^
j
√
2
cos
(
k
z
+
ω
t
)
At
t
=
0
, a positively charged particle is at the point
(
x
,
y
,
z
)
=
(
0
,
0
,
π
k
)
. If its instantaneous velocity at
(
t
=
0
)
is
υ
0
^
k
, the force acting on it due to the wave is :
zero
parallel to
^
i
+
^
j
√
2
antiparallel to
^
i
+
^
j
√
2
parallel to
^
k
Validate
Solution:
→
F
=
q
(
→
E
+
→
v
×
→
B
)
→
E
=
E
0
(
^
i
+
^
j
√
2
)
cos
π
=
−
E
0
(
^
i
+
^
j
√
2
)
as
→
E
×
→
B
=
→
c
+
E
0
(
^
i
+
^
j
√
2
)
×
→
B
=
→
c
^
k
⇒
→
B
=
−
(
^
i
−
^
j
√
2
)
E
0
c
→
F
=
q
(
−
E
0
(
^
i
+
^
j
√
2
)
−
v
0
^
k
c
×
(
^
i
+
^
j
)
E
0
)
Since
v
0
c
<
<
1
⇒ F is antiparallel to
^
i
+
^
j
√
2
© examsnet.com
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