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.

Explanation

Energy o fthe ion, `E = q xx V = 2 xx 10 ^(-16) xx 200 J = 400 xx 10 ^(-16) J`
`B = 2 xx 10 ^(-3) T, m = 1. 8 xx 10 ^(-27) kg`
As, `E . = (1)/(2) mv ^(2) implies v = sqrt (( 2 E)/( m )). Also ` bqv `= ( mv ^(2))/( r)`
or `r = ( mv )/(Bq ) = ( m )/( Bq) sqrt (( 2 E .)/( m )) = (sqrt ( 2 E.m))/( Bq)or , r = (sqrt (2 xx 400 xx 10 ^(-16) 1. 8 xx 10 ^(-27)))/( 2 xx 10 ^(-3) xx 2 xx 10 ^(-16)) = 3 xx 10 ^(-2) m`