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In a region `10^(19) prop`-particles and `10^(19)` protons move to the left, while `10^(19)` electrons moves to the right per second. The current isA. 3.2 A towards leftB. 6.4 A towards leftC. 9.6 A towards leftD. 6.4 A towards right |
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Answer» Correct Answer - B Electric current due to flow of `10^(19) alpha-`particles towards left is , `I_(1)=2 n_(1)e=2xx10^(19)xxe` Electric current due to flow of `10^(19)` electrons towards left is , `I_(2)=10^(19)xxe` Electric current due to flow of `10^(19)` protons towards left is , `I_(3)=10^(19)xxe` Thus, total electric current is , `I=I_(1)+I_(2)+I_(3)=4e xx 10^(19)` `=4xx1.6 xx 10^(-19)xx10^(19)=6.4` A |
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