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An electron travels in a circular path of radius `20cm` in a magnetic field `2xx10^-3T` (i) Calculate the speed of the electron (ii) What is the potential difference through which the electron must be accelerated to acquire this speed? Charge of electron `=1*6xx10^(-19)C`. Mass of electron `=9xx10^(-31)kg`. |
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Answer» Here `r=20xx10^-2m`, `B=2xx10^-3T`, (i) Magnetic force on the electron=centripetal force or `evB=(mv^2)/(r)` or `v=(eBr)/(m)` `=((1*6xx10^(-19))xx(2xx10^-3)xx(20xx10^2))/(9xx10^(-31))` `=7*1xx10^7ms^-1` (ii) Let V be the pot. diff. required to provide speed v to the electron, then `eV=1/2mv^2` or `V=(mv^2)/(2e)=(9xx10^(-31)xx(7*1xx10^7)^2)/(2xx(1*6xx10^(-19))` `=14*2xx10^3V=14*2kV` |
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