CBSE Class 12-science Answered
Consider a particle of mass m and charge q entering a uniform magnetic field O with velocity v, making an angle θ with the direction of magnetic field.
Resolving into two rectangular components:
i) v1= vcosθ along the direction of .
ii) v2=vsinθ perpendicular to .
Force acting on the charged particle due to v1
1= q(1 x )
F1= qvBsin0° [Because the angle between 1 & is zero]
Therefore F1= 0.
Thus, the particle covers the linear distance in the direction of magnetic field.
Force acting on the charged particle due to v2
2= q (2 x )
F2= q v2B sin90°
F2= q v2B
F2= q (vsinθ) B [Since v2= vsinθ]
The direction of this force is perpendicular to 2. So, it cannot change the magnetic magnitude of velocity 2 but changes only the direction of motion. Hence, the charged particle is made to move on the circular path in the magnetic field as shown in the figure. Thus, the force F2 provides the required centripetal force {= (mv22)/r} necessary for motion along the circular path of radius r.
Therefore q v2B= {(mv22)/r}
v2= (Bqr)/m [as v2=vsinθ ]
The velocity of rotational
The frequency of rotation
The time period of revolution
Thus, we conclude that under the combined effect of two components velocities, the charged particle in magnetic field will cover linear path as well as circular path i.e. the path will be helical, whose axis is parallel to the direction of the magnetic field.