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An electron emitted by a heated cathode and accelerted through a potential difference of `2.Kv,` enters a region of uniform magnetic field of `0.15T`. Determine the trajectory of the electron if the field makes an angle `30^(@)` with the initial velocity. |
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Answer» When magnetic field makes an anlge `30^(@)` with the initial velocity i.e. `theta=30^(@)` Then `V^(1)=V sin theta =sqrt((2ev)/(m))=sin 30^(@)` `V^(1)=(8)/(3)xx10^(7)xx(1)/(2)=(4)/(3)xx10^(7) m//s` The radius of the helical path is `r=(mV^(1))/(Be)` `=(9xx10^(-31)xx((4)/(3)xx10^(7)))/(0.15xx1.6xx10^(-19))=0.5 mm` |
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