InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A charged particle moving in a uniform magnetic field penetrates a layer of lead and there by loses one-half of its kinetic energy. How does the radius of curvature of its path change?A. The radius increases to `r sqrt(2)`B. The radius reduces to `r//sqrt(2)`C. The radius remains the sameD. The radius becomes `r//2` |
|
Answer» Correct Answer - B `E_2 =(E_1)/(2), r_1=r,r_2 =?` `E=(q^2 B^2 r^2)/(2m)` `therefore E prop r^2` `therefore r prop sqrt( E)` `(r_2)/(r_1)=sqrt((E_2)/(E_1))` `therefore r_2=sqrt((E_1)/(2E_1))r_1 =(r_1)/(sqrt(2))= (r ) /(sqrt(2))`. |
|