Question

A deutron of kinetic energy 50 keV is describing a circular
orbit of radius 0.5m, in a plane perpendicular to magnetic
field →B. The kinetic energy of a proton that discribes a
circular orbit of radius 0.5m in the same plane with the same
magnetic field →B is
(a) 200 keV (b) 50 keV (c) 100 keV (d) 25 keV

Answer 1

For a charged particle orbiting in a circular path in a magnetic field

r

mv

2

=Bqv⇒v=

m

Bqr

or, mv

2

=Bqvr

Also,

E

K

=

2

1

mv

2

=

2

1

Bqvr=Bq

2

r

.

m

Bqr

=

2m

B

2

q

2

r

2

For deuteron, E

1

=

2×2m

B

2

q

2

r

2

For proton, E

2

=

2m

B

2

q

2

r

2

E

2

E

1

=

2

1

⇒

E

2

50keV

=

2

1

⇒E

2

=100keV.

Answer 2

Answer:

A is the right answer.

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