Q.

A charged particle moves in a uniform magnetic field perpendicular to it,   in a  circle of radius of  4 cm. On passing through a metallic sheet it loses half of its kinetic energy. Then, the radius of curvature of the particle is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

2 cm

b

4 cm

c

8 cm

d

22 cm

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

R=mvqB=mqB2(KE)m=2mqBKE⇒  R1R2=KEKE/2⇒R2=R12=42=22 cm
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
A charged particle moves in a uniform magnetic field perpendicular to it,   in a  circle of radius of  4 cm. On passing through a metallic sheet it loses half of its kinetic energy. Then, the radius of curvature of the particle is