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VITEEE 2015 Solved Paper
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© examsnet.com
Question : 19
Total: 120
An electron of mass 9.0 x
10
−
31
kg under the action of a magnetic field moves in a circle of radius 2 cm at a speed of 3 x
10
6
m/s. If a proton of mass 1.8 x
10
27
kg was to move in a circle of same radius in the same magnetic field, then its speed will become
1.5 x
10
3
m/s
3 x
10
6
m/s
6 x
10
4
m/s
2 x
10
6
m/s
Validate
Solution:
Here, the magnetic force (Bqv) will provide the necessary centripetal force
(
m
v
2
r
)
∴ Bqv =
m
v
2
r
⇒ Bqv =
m
v
2
r
⇒ Bqv = mv
For electron and proton, the magnetic field B, charge q and radius r, all same
∴ mv = constant
i.e.
m
e
v
e
=
m
p
v
p
v
p
=
(
m
e
m
p
)
v
e
=
(
9
×
10
−
31
1.8
×
10
−
27
)
3 ×
10
6
= 1.5 ×
10
3
m/s
© examsnet.com
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