InterviewSolution
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InterviewSolution
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A beam of alpha particle passes underflected with a horizontal velocity v, through a region of electric and magnetic fields, mutually perpendicular to each other and normal to the direction of the beam. If the mangnitude of the electric and magnetic fields are `120kVm^-1` and `60mT` respectively, calculate (i) velocity of the beam (ii) force with which it strikes a target on a screen, if the alpha particle beam current is equal to `1mA`? Mass of proton=mass of neutron `=1*675xx10^(-27)kg`. |
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Answer» Here, `E=120kVm^-1` `=120xx10^3Vm^-1`, `B=60mT=60xx10^-3T` (i) For undeflected beam of alpha particle `qvB=qE` or `v=E/B=(120xx10^3)/(60xx10^-3)` `=2xx10^6ms^-1` (ii) Current carried by alpha particle beam `I=1mA=10^-3A` Number of alpha particles striking the screen per second `n=I/q=(I)/(2e)=(10^-3)/(2xx1*6xx10^(-19))` `=3*125xx10^(15)s^-1` Mass of alpha particle, `m_alpha=2m_p+2m_n=4m_p` `=4xx1*675xx10^(-27)kg` Force with which alpha particle beam strikes a target on the screen `F=m_alphanv=(4xx1*675xx10^(-27))` `xx(3*125xx10^(15))xx2xx10^6` `=4*19xx10^-5N` |
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