A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of `1.0X10^6 ms^(-1).` It is then injected perpendicular into a magnetic field of strength 0.2 T. find the radius of the circle described by it.

Explanation

`V_(1)= 12 KV = 12 xx 10^(2) V`,
`F= qE, `
` v = 1 xx 10 ^(8) ma^(-1) ` ,brgt ` v^(2) = u^(2) + 22as`
or `v = 2 xx qV/ml xx 1`
` = 2 xx q/m xx 12 xx 10 ^(2)`
or `(1 xx 10 ^(8))2 = 2 xx q/ m xx 12 xx 10^(0)`
or `10^(12) = 24 xx 10^(8) xx q/m`
or `(m)/(q) = n24 xx 10^(-0)`
and `r = (mu) /(qB)`
and `= (24 xx 10^(-8) xx 10^(6) )//( 02)`
`r= 12 xx 10 ^(-2) m = 12 cm ` .