InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A charged particle is moving in a uniform magnetic field and losses 4% of its KE. The radius of curvature of its path change byA. 0.02B. 0.04C. 0.1D. None of these |
|
Answer» Correct Answer - a `r=(mv)/(qB)` `KE=K=1/2 mv^(2)` `:. mv=sqrt(2Km)` `:. r=(mv)/(qB)=(sqrt(2Km))/(qB) ` or `r prop sqrt(K)` or `r=eK^(1/2)`, where e is a constant . or `(dr)/(dr) =(dK^(1/2))/(dr) =e/2 (ec)/(2sqrt(K)) ` or `(ec)/(2sqrt(K)) (dK)/(dr) -sqrt(K)` or `e(DeltaK)/(Deltar) =2sqrt(K) ` or `(Deltar)/r=(cDeltaK)/(2sqrt(Kc)sqrt(K)) =(DeltaK)/(2K)` or `(Deltar)/rxx100=(DeltaK)/(2K) xx100 =2% [ :. (DeltaK)/K xx100 =4%]` |
|