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A particle of mass `1.6xx10^(-27) kg` and charge `1.6 xx 10^(-19)` coulomb enters a uniform magnetic field of 1 Tesla as shown in the figure. The speed of the particle is `10^7 m//s`. The distance PQ will be A. 0.14 mB. 0.28 mC. 0.4 mD. 0.5 m |
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Answer» Correct Answer - A `PQ=2r sin theta=2(mv)/(qB) sin theta=(2xx1.6xx10^(-27)xxsin45)/(1.6xx10^(-19)xx1)=2/(10sqrt(2)) =(sqrt(2))/10=0.14` |
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