A Proton moves in a iniform magnetic field of strength 0.500 T magnetic field is directed along the x-axis . At initial time, t = 0 s, the proton has velocity `vec(v) = (1.95 xx 10^(5) hat(i) + 2.00 xx 10^(5) hat(k) ) ms^(-1)`. Find
(a) At initial time, what is the acceleration of the proton
(b) is the path circular or helical , calculate the radius of helical trajectory and also calculate the pitch of the helix (Note: Pitch of the helix is the distance travelled along the helix axis per revolution ).

Explanation

Magnetic field `vec(B) = 0.500 hat(i) ` T
Velocity of the particle `vec(v) = (1.95 xx 10^(5) hat(i) + 2.00 xx 10^(5) hat(k) ) mas^(-1)`
Charge of the proton q = 1.60 `xx 10^(-19)` C
Mass of the proton m = `1.67 xx 10^(-27) `kg
(a) the force experienced by the proton is `vec(F) q (vec(v) xx vec(B) )`
= ` 1.60 xx 10^(-19) xx ((1.95 xx 10^(5) hat(i) + 2.00 xx 10^(5) hat(k)) xx (0.500 hat(i)) ) `
Therefore, from Newton.s second law,
`vec(a) = (l)/(m) vec(F) = (l)/(1.67 xx 10^(-27)) (1.60 xx 10^(-14)) `
= 9.58 `xx 10^(12) ms^(-2)`
(b) Trajectory is helical
Radius of helical path is
`R = (mv_(z))/(|q|B) = (1.67 xx 10^(-27) xx 2.00 xx 10^(5) )/(1.60 xx 10^(-19) xx 0.500) = 4.175 xx 13^(-3) m = 4.18` mm
Pitch of the helix is the distance travelled along x-axis in a time T, which is P = `v_(x)` T But time,
`T = (2pi)/(omega) = (2pi m)/(|q|B ) = (2xx3.14 xx 1.67 xx 10^(-27) )/(1.60 xx 10^(-19) xx 0.500) = 13.1 xx 10^(-8) `s
hence , pitch of the helix is
`P = v_(x) T = (1.95 xx 10^(4)) (13.1 xx 10^(-8)) = 25.5 xx 10^(-3) m = 25.5 `mm
The proton experiences appreciable acceleration in the magnetic field, hence the pitch of the helix is almost six times greater than the radius of the helix.