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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

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`m = 1*8 xx 10^(-27)` kg
`q = 2 xx 10^(16)` C
`v = 200` volt, `B = 2 xx 10^(-3)` T
Radius = r = ?
`r = (mv)/(Bq)`
But `(1)/(2) mv^(2) = qV`
`v = sqrt((2qV)/(m))`
`v = sqrt((2 xx 2 xx 10^(-6) xx 200)/(1*8 xx 10^(-27)))`
`v = sqrt(4*44 xx 10^(13))`
`v = sqrt(44*4 xx 10^(12))`
`v = 6*67 xx 10^(-6)ms^(-1)`
`therefore r = (1*8 xx 10^(-27) xx 6*67 xx 10^(6))/(2 xx 10^(-3) xx 2 xx 10^(-6))`
`r = 3*0 xx 10^(-2)` m
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