Saved Bookmarks
| 1. |
Two particles of masses M_(1)andM_(2) and having the equal electric charge are accelerated through equal potential difference and then move inside a uniform magnetic field, normal to it. If the radii of their circular paths are R_(1)andR_(2) respectively, find the ratio of their masses. |
|
Answer» Solution :1. Two particles having mass `M_(1)andM_(2)` and equal electric CHARGE move under potential v (suppose) 2. For a particle having mass `M_(1)=qV=1/2M_(1)v_(1)^(2)` For a particle having mass `M_(2)=qV=1/2M_(2)v_(2)^(2)` `therefore1/2M_(1)v_(1)^(2)=1/2M_(2)v_(2)^(2)` `thereforeM_(1)/M_(2)=v_(2)^(2)/v_(1)^(2)""...(1)` 3. After ENTERING into uniform magnetic field, both the particles perform uniform circular motion DUE to CENTRIPETAL FORCE acting on them. For a particle having mass `M_(1)=(M_(1)v_(1)^(2))/R_(1)=qv_(1)B` `therefore(M_(1)v_(1)^(2))/R_(1)=qB` Similarly for a particle having mass `M_(2)`, `(M_(2)v_(2)^(2))/R_(2)=qv_(2)B` `therefore(M_(2)v_(2))/R_(2)=qB` `therefore(M_(1)v_(1))/R_(1)=(M_(2)v_(2))/R_(2)` `thereforeM_(1)/M_(2)=v_(2)/v_(1)R_(1)/R_(2)` From equation (1) `v_(2)/v_(1)=(M_(1)/M_(2))^(1/2)` `thereforeM_(1)/M_(2)=(M_(1)/M_(2))^(1/2)*R_(1)/R_(2)` `therefore(M_(1)/M_(2))^(1/2)=R_(1)/R_(2)` `thereforeM_(1)/M_(2)=(R_(1)/R_(2))^(2)` |
|