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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 ). |
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Answer» Solution :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. |
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