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The momentum of `alpha`-particles moving in a circular path of radius 10 cm in a perpendicular magnetic field of 0.05 tesla will be :A. `1.6xx10^(-20)km` m/sB. `1.6xx10^(-21)kb` m/sC. `1.6xx10^(-19)kg` m/sD. `1.6xx10^(-18)kg` m/s |
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Answer» Correct Answer - B `qvB=(mv^(2))/(R)` `mv=qRB=1.6xx10^(-21)kg` m/sec |
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