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Q.
A proton is projected with a uniform velocity $v$ along the axis of a current-carrying solenoid, then
NTA AbhyasNTA Abhyas 2022
Solution:
The force on a charged particle in magnitude field
$F=q\left(v \times B\right)$
The proton is moving along the axis of solenoid. Magnetic field inside the solenoid is parallel to the axis of solenoid.
$\therefore \overset{ \rightarrow }{v} \parallel \overset{ \rightarrow }{B}$
$F=qvBsin \theta $
$\theta =0$
$F=0$
$\therefore \, \, $ The proton will continue to move with velocity along the axis.