InterviewSolution
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InterviewSolution
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The electron in a hydrogen atom circles around the proton with a speed of `2*18xx10^6ms^-1` in an orbit of radius `5*3xx10^-11m`. Calcualte (a) the equivalent current (b) magnetic field produced at the proton. Given charge on electron `1*6xx10^-19C` and `mu_0=4pixx10^-7TmA^-1`. |
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Answer» Here, `v=2*18xx10^6ms^-1`, `r=5*3xx10^-11m`, `e=1*6xx10^-19C`. (a) Time period of revolution of electron is given by, `T=(2pir)/(v)=(2pixx5*3xx10^-11)/(2*18xx10^6)=1*528xx10^-16s` Equivalent current, `I=(charg e)/(time)=e/T` `=(1*6xx10^-19)/(1*528xx10^-16)=1*05xx10^-3A` (b) `B=(mu_0)/(4pi)(2piI)/(r)=(10^-7xx2pixx1*05xx10^-3)/(5*3xx10^-11)` `=12*4T` |
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