A particle of charge q and mass m starts moving from the origin under the action of an electric field →E=E0^i and →B=B0^i with velocity →v=v0^j. The speed of the particle will become 2v0 after a time
Answers
Answered by
0
Correct option is
D
t=
qE
3
mv
0
E
is parallel to
B
and
v
is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t is
v=
v
x
2
+v
y
2
+v
z
2
....(1)
Here, v
y
2
+v
z
2
=v
0
2
And v
x
2
=(
m
qE
t)
2
And v=2v
0
Substituting the value in eq (1), we get
t=
qE
3
mv
0
Similar questions