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An ion of mass 1. 8 xx 10 ^(-27) kg car...

An ion of mass `1. 8 xx 10 ^(-27) ` kg carrying a charge `2 xx 10 ^(- 16) C` after being accelerated through a potential difference of 200 V enters a uniform magnetic field of intensity `2 xx 10 ^(-3) T bot r ` to the direction of motion. Calculate the radius of the path described by the ion.

Text Solution

Verified by Experts

`m = 1*8 xx 10^(-25)` Kg
`q = 4 xx 10^(-16)` C
` V = 400` volt
`B = 2 xx 10^(-2)` T
Radius = r = ?
`(1)/(2)mv^(2) = qV`
`v = sqrt((2qV)/(m))`
`v = sqrt((2xx 4 xx 10^(-16) xx 400)/(1*8 xx 10^(-25)))`
` = sqrt(17*78 xx 10^(11))`
`v = sqrt(177*8 xx 10^(10))`
` = 13*3 xx 10^(5)ms^(-1)`
Now `r = (mv)/(Bq)`
`r = (1*8 xx 10^(-25) xx 13*3 xx 10^(5))/(2xx10^(-2) xx 4 xx 10^(-16))`
` = 2*99 xx 10^(-2)`
` = 3*0 xx 10^(-2)` m
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