KCET 2018 Physics Solved Paper

We know, radius of circular path
r=‌

mv

qB

⇒v=‌

qBr

m

→ (1)
Frequency of cyclotron, f=

qB

2πm

⇒B=

2πmf

q

→ (2)
Using Eq. (1), kinetic energy =

1

2

mv2=

1

2

m

q2B2r2

m2

Using Eq. (2), kinetic energy =‌

1

2

‌

q2r2

m

‌

(2π)2m2f2

q2

=2π2f2r2m
Therefore, kinetic energy =2×(3.14)2×(10×106)2× (60×10−2)2×(1.67×10−27)
=1.13×10−12J
Hence, kinetic energy