Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

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

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

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
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
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.