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JEE Advanced Model Paper 5 with solutions for online practice
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© examsnet.com
Question : 4
Total: 54
A charged particle having specific charge β moves as a conical pendulum of length l making an angle θ with thevertical and maintaining a constant angular velocity ω, in the presence of a uniform magnetic field B is directedvertically downward as shown in figure, then
Magnitude of magnetic field is
1
β
(
ω
−
g
ω
l
c
o
s
θ
)
Magnitude of angular momentum of the particle about the point of suspension remains constant
Magnitude of magnetic field is
1
β
(
ω
+
g
ω
l
c
o
s
θ
)
Rate of change of angular momentum of the particle about the axis of rotation is not a constant vector
Validate
Solution:
Answer (A, B)
T cos θ = mg
T sinθ + qωl sinθ B =
m
ω
2
lsinθ
B =
1
β
(
ω
−
g
ω
l
c
o
s
θ
)
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