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An electron having momentum 2.4 xx 10^(-23)" kg m s"^(-1) enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30^(@) with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be |
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Answer» 2mm `r=(mv_(_|_))/(eB)=("mv sin "theta)/(eB)=((2.4xx10^(-23)"kg m s"^(-1))xxsin 30^(@))/((1.6xx10^(-19)C)xx(0.15T))` `=5xx10^(-4)m=0.5xx10^(-3)m=0.5mm` |
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