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Cross-sectional area of proton beam having electric current 1 muA is 0.5mm^(2) and move with velocity 3 xx 10^(4) m/s so current density = ......... |
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Answer» `6.6 xx 10^(-4) C // m^(3)` Electric charge density `rho= (" electric charge ")/(" Volume ") = (Q)/(AD)` ` therefore rho =(It)/(Ad) = (I)/(A xx v) "" [ because (d)/(t) = v ] ` `therefore rho =(10^(-6))/(0.5 xx 10^(-6) xx 3 xx 10^(4))` `therefore rho = 0.66 xx 10^(-4) C// m^(3)` `therefore rho = 6.6 xx 10^(-5) C//m^(3)` |
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