Q.

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of α-particle (mass 4 m and charge + 2 e) so that it can revolve in the path of same radius

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

1 MeV

b

4 MeV

c

2 MeV

d

0 .5 MeV

answer is A.

(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

we know that r=mvqB  or  r2=m2v2q2B2In terms of energy, the above equation can be written asr2=2m×1/2mv2q2B2=2mEq2B2Here r and B are same, hence    m1E1q12=m2E2q22 or m×1e2=3m×E24e2∴     E2=1MeV
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