COMEDK Exam 2014 Solved Paper

Radius of circular path of a charged particle in uniform magnetic field, when it enters perpendicular direction of magnetic field.
r=

mv

Bq

................(i)
where, m= mass, v= velocity and q= charge.
We know that, kinetic energy,
K=

1

2

mv2
⇒v=√

2K

m

....................(ii)
From Eqs. (i) and (ii), we get
r=

m

Bq

.√

2Km

Bq

For the same value of kinetic energy,
r∝√

m

q

⇒

rp

rα

=√

mp

mα

.

qα

qp

=√

mp

4mp

.(

2qp

qp

)=

1

1

‌or‌1:1