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A mono energetic electron beam with a speed of ` 5.2xx10^(6) ms^(-1)` enters into a magnetic filed of induction ` 3xx10^(-4) T`, directed normal to the beam . Find the radius of the circul traced by the beam `(Take e//m=1.76xx10^(11) Ckg)` |
Answer» From the eqation, `Bev =(mv^(2))/(r)` Radius of circle `=(mv)/(Bq)= V/(B(e//m))` `therefore r=(5.2xx10^(6))/(3xx10^(-4) (1.76xx10^(11)))=01.m` |
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