A charged particle enters a uniform magnetic field with velocity v0=4 m/s perpendicular to it, the length of magnetic field is x=32R, where R is the radius of the circular path of the particle in the field. Find the magnitude of change in velocity (in m/s ) of the particle when it comes out of the field.
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 4.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The particle will come out of the magnetic field at an angle θ=600with the original direction.Δv→=v0cos60i^+v0sin60j^-v0i^=-v022+v0322 ⇒|Δv→|=v0
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
A charged particle enters a uniform magnetic field with velocity v0=4 m/s perpendicular to it, the length of magnetic field is x=32R, where R is the radius of the circular path of the particle in the field. Find the magnitude of change in velocity (in m/s ) of the particle when it comes out of the field.