A particle of mass m and charge q moving with velocity v=qBl2m enters region-II normal to the boundary as shown in the figure. Region-II has a uniform magnetic field B perpendicular to the plane of paper. The particle will
A
Enter the region - III
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Not enter the region-III
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Complete circular motion in region-II
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Move undeflected in region-II
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A Not enter the region-III Given :A particle of mass m and charge q moving with velocity v=qBl2mv=qBl2m enters region-II normal to the boundary as shown in the figure. Region-II has a uniform magnetic field B perpendicular to the plane of paper.
Solution :
Formula for radius of circular path when a charge enters magnetic field is R=mvqB
And we have the velocity of charge V= qBl2m
when we substitute this value in the formula of radius we get R= l2
It means particle will move in semicircular path and come out with deviation of 180