(a) A charged particle having mass `m` and charge `q` is accelerated by a potential difference `V`, it flies through a uniform transverse magnetic field `B`. The field occupies a region of space `d`. Find the time interval for which it remains inside the magnetic field.
(b) An `alpha`-particle is acceleration by a potential difference of `10^(4) V`. Find the change in its direction of motion if it enters normally in a region of thickness `0.1 m` having transverse magnetic induction of `0.1 T`.
`(m_(alpha) = 6.4 xx 10^(-27) kg)`.
( c) A `10 g` bullet having a charge of `4 muC` is fired at speed of `270 m//sec` in a horizontal direction. A vertical magnetic field of `500 muT` exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through `100 m`. Make appropriate approximations.

Explanation

(a) `K = (1)/(2)mv^(2) - qV`
`v = sqrt((2qV)/(m))`
`sin theta = (d)/(R ) = (d)/(R ) = (d)/(mv// Bq) = (Bqd)/(mv)`
(b) `V = 10^(4)V, q = 2e = 2 xx 1.6 xx 10^(-19)C`,
`d = 0.1m, B = 0.1 T, m_(alpha) = 6.4 xx 10^(-27) kg`
`K = (1)/(2)m_(alpha)v^(2) = qV = 2 eV`
`v = sqrt((4 eV)/(m_(alpha)))`
`R = (m_(alpha)v)/(Bq) = (m_(alpha)v)/(2eB)`
`sin theta = (d)/(R ) = (d)/(m_(alpha)v//B.2e) = (2eBd)/(m_(alpha)v)`
`= (2eBd)/(m_(alpha)sqrt((4eV)/(m_(alpha)))) = sqrt((e)/(m_alphaV)).Bd`
`= sqrt((1.6 xx 10^(-19))/(6.4 xx 10^(-27) xx 10)) xx 0.1 xx 0.1`
`= (1)/(2)`
`theta = 30^(@)`
( c) `m = 10 g = 10 xx 10^(3) = 10^(-2) kg`
`q = 4 muC = 4 xx 10^(-6)C`
`v = 270 m//sec`
`B = 500 muT = 500 xx 10^(-6) = 5 xx 10^(-4)T`
`d = 100 m`
`R = (mv)/(Bq) = (10^(-2) xx 270)/(5 xx 10^(-4) xx 4 xx 10^(-6)) = 13.5 xx 10^(8) m`
`sin theta = (d)/(R ) = (100)/(13.5 xx 10^(8)) = (10^(-6))/(13.5)`
`theta` is very small.
Deflection `= (d^(2))/(2R)` (as proved in previous example) `= ((100)^(2))/(2 xx 13.5 xx 10^(8)) = 3.7 xx 10^(-6) m`