A proton moves in a uniform magnetic field of strength 0.500 T magnetic field is directed along the x-axis. At initial time, t s = 0 , the proton has velocity `vec(v) = (1.95 xx 10^(5)hat(i) + 2.00 xx 10^(5) hat(k)) ms^(-1)`. Find
Is the path circular or helical?. If 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))ms^(-1)`
Charge of the proton `q = 1.60 xx 10^(-19)C`.
Mass of the proton `m = 1.67 xx 10^(-27) kg`
Trajectory is helical
Radius of helical path is
`R = (m v_(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 10^(-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) = (2 xx 3.14 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^(5))(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.