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A chamber is maintained a uniform magnetic field of `5xx10^-3T`. An electron with a speed of `5xx10^7ms^-1` enters the chamber in a direction normal to the field. Calculate (i) radius of the path (ii) frequency of revolution of the electron. Charge of electron `=1*6xx10^(-19)C`, Mass of electron `=9*1xx10^(-31)kg` |
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Answer» Correct Answer - (i) `5*7cm` (ii) `1*4xx10^8Hz` Here, `B=5xx10^-3T`, `v=5xx10^7ms^-1`, `e=1*6xx10^(-19)C`, `m=9*1xx10^(-31)kg` (i) Radius of the circular path, `r=(mv)/(Be)=((9*1xx10^(-31))xx(5xx10^7))/((5xx10^-3)xx(1*6xx10^(-19)))` `=5*7xx10^-2m` `=5*7cm` (ii) Frequency of revolution, `v=(eB)/(2pim)=((1*6xx10^(-19))xx(5xx10^-3))/(2xx3*14xx(9*1xx10^(-31))` `=1*4xx10^8Hz` |
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