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An electron of mass `0*90xx10^-30kg`, under the action of magnetic field moves in a circle of radius `2*0cm` at a speed of `3*0xx10^6ms^-1`. If a proton of mass `1*8xx10^-27kg` were to move in a cirlce of the same radius in the same magnetic field, find its speed. |
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Answer» Here, `m_e=9xx10^-31kg`, `r_e=0*02m`, `v_e=3xx10^6ms^-1`, `m_p=1*8xx10^-27kg`, `v_p=?`, `r_p=0*02m`. We know that `qvB=mv^2//r` or `v=qrB//m`, So `v_p/v_e=(q_pr_pB//m_p)/(q_er_eB//m_e)=(m_e)/(m_p)` [as `q_p=q_e`, `r_p=r_e`] `:. v_p=v_exxm_e/m_p=(3xx10^6xx9xx10^-31)/(1*8xx10^-27)=1*5xx10^3ms^-1` |
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