A uniform magnetic field `B = 1.2` mT is directed vertically upward throughout the volume of a laboratory chamber . A proton (`m_(p) = 1.67 xx 10^(-27)` kg) enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is `3 xx 10^(7)` m/s

Options

A. `3.45 xx 10^(12) m//s^(2)`
B. `1.67 xx 10^(12) m//s^(2)`
C. `5.25 xx 10^(12) m//s^(2)`
D. `2.75 xx 10^(12) m//s^(2)`

Answer

A

Explanation

`B = 1.2 xx 10^(-3) (hatk)` T
`v = 3 xx 10^(7) (hatj)` m/s
`r = (mv)/(qB) = (1.67 xx 10^(-27) xx 3 xx 10^(7))/(1.6 xx 10^(-19) xx 1.2 xx 10^(-3))`
`= 2.6 xx 10^(2)` m
Centripetal acceleration - `a_(c) = (v^2)/( r) = ((3 xx 10^(7))^(2))/(2.6 xx 10^(2)) = 3.46 xx 10^(12) m//s^(2) = 3.45 xx 10^(12) m//s^(2)`