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JEE Mains 02-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 1
Total: 75
PHYSICS
The figure shows a region of length '
l
with a uniform magnetic field of
0.3
T
in it and a proton entering the region with velocity
4
×
10
5
m
s
−
1
making an angle
60
°
with the field. If the proton completes 10 revolution by thetime it cross the region shown, '
l
′
is close to (mass of proton
=
1.67
×
10
−
27
k
g
,
charge of the proton
=
1.6
×
10
−
19
C
)
0.11
m
0.22
m
0.44
m
0.88
m
Validate
Solution:
T
=
2
π
m
q
B
total time
t
=
10
T
Kinematics
l
=
V
2
t
l
=
V
2
10
×
2
π
m
q
B
=
4
×
10
5
×
10
×
3.14
×
1.67
×
10
−
27
1.6
×
10
−
19
×
0.3
=
0.439
© examsnet.com
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