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JEE Main Physics Class 12 Electric Charges and Fields Questions
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© examsnet.com
Question : 48
Total: 108
A charged particle of mass '
m
' and charge 'q' moving under the influence of uniform electric field
E
^
i
and a uniform magnetic field
B
→
k
follows a trajectory from point
P
to
Q
as shown in figure. The velocities at
P
and
Q
are respectively,
v
→
i
and
−
2
v
→
j
. Then which of the following statements
(
A
,
B
,
C
,
D
)
are the correct? (Trajectory shown is schematic and not to scale)
(A)
E
=
3
4
(
m
v
2
q
a
)
(B) Rate of work done by the electric field at
P
is
3
4
(
m
v
2
a
)
(C) Rate of work done by both the fields at
Q
is zero
(D) The difference between the magnitude of angular momentum of the particle at
P
and
Q
is 2 mav.
[9 Jan. 2020, I]
(A), (C), (D)
(B), (C), (D)
(A), (B), (C)
(A), (B), (C), (D)
Validate
Solution:
(A) By work energy theorem
W
m
g
+
W
e
l
e
=
1
2
m
(
2
v
)
2
−
1
2
m
(
v
)
2
0
+
q
E
0
2
a
=
3
2
m
v
2
⇒
E
0
=
3
4
m
v
2
q
a
(B) Rate of work done at
P
=
power of electric force
=
q
E
0
V
=
3
4
m
v
3
a
(C)
A
t
,
Q
,
d
w
d
t
=
0
for both the fields
(D) The difference of magnitude of angular momentum of the particle at P and Q,
Δ
→
L
=
(
−
m
2
v
2
a
^
k
)
−
(
−
m
v
a
^
k
)
|
Δ
→
L
|
=
3
m
v
a
© examsnet.com
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