The electric field of a plane electromagnetic wave is given by →E=E0 ^i+^j√2cos(kz+ωt) At t=0, a positively charged particle is at the point (x,y,z)=(0,0,πk). If its instantaneous velocity at (t=0) is v0^k, the force acting on it due to the wave is: a.parallel to ^i+^j√2 b.zero c.antiparallel to ^i+^j√2 d.parallel to ^k
Answers 1
The correct option is C antiparallel to
^
i
+
^
j
√
2
At
t
=
0
,
z
=
π
k
⇒
→
E
=
E
0
√
2
(
^
i
+
^
j
)
cos
[
π
]
⇒
→
E
=
−
E
0
√
2
(
^
i
+
^
j
)
The force on the charged particle
q
due to Electric field is given by,
−→
F
E
=
q
→
E
Therefore, force due to electric field is in direction
−
(
^
i
+
^
j
)
√
2
.
In an electromagnetic wave, the magnetic field exists in the direction perpendicular to both the electric field and the direction of propagation.
Therefore, force due to magnetic field is in direction
−→
F
B
=
q
(
→
v
×
→
B
)
and here
→
v
|
|
→
k
Therefore,
−→
F
B
is parallel to
→
E
.
⇒
→
F
n
e
t
=
−→
F
E
+
−→
F
B
As both
−→
F
E
and
−→
F
B
are parallel to
→
E
,
⇒
→
F
n
e
t
is antiparallel to
^
i
+
^
j
√
2
Hence, option
(C)
is correct
-
Author:
gabrielharris
-
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