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BITSAT 2016 Solved Question Paper
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© examsnet.com
Question : 14
Total: 150
A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is
25 keV
50 keV
200 keV
100 keV
Validate
Solution:
For a charged particle orbiting in a circular path in a magnetic field
m
v
2
r
=
B
q
v
⇒
v
=
B
q
r
m
or
m
v
2
=
B
q
v
r
Also
E
K
=
1
2
m
v
2
=
1
2
B
q
v
r
=
B
q
r
2
⋅
B
q
r
m
=
B
2
q
2
r
2
2
m
For deuteron,
E
1
=
B
2
q
2
r
2
2
×
2
m
For proton,
E
2
=
B
2
q
2
r
2
2
m
E
1
E
2
=
1
2
⇒
50
k
e
V
E
2
=
1
2
⇒
E
2
=
100
k
e
V
© examsnet.com
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