Cross-sectional area of proton beam having electric current 1 `mu`A is 0.5 `mm^(2)` and move with velocity `3 xx 10^(4)` m/s so current density = .........

Options

A. `6.6 xx 10^(-4) C // m^(3)`
B. `6.6 xx 10^(-5) C// m^(3)`
C. `6.6 xx 10^(-6) C// m^(3)`
D. None of these

Answer

B

Explanation

`6.6 xx 10^(-5) 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)`