KCET 2021 Physics Solved Paper

Magnetic force on the charged particle is given as
FB=q(v×B)=qvB‌sin‌θ
Here ,θ=90°
⇒ FB=qvB‌sin‌90°=qvB
Since, the particle is projected perpendicular to the field, so its trajectory will be circular in nature, where necessary centripetal force =FB
⇒

mv2

r′

=qvB

mv2

( r 2 )

=qvB (∵r′=

r

2

)
⇒ v=

qBr

2n

...(i)
Magnetic field inside the solenoid is given as,
B=µ0‌nI
Substituting the value of B in Eq. (i), we get
v=

µ0‌nIqr

2m