A beam of protons enters a uniform magnetic field of 0.3 t0.3 t with a velocity of 4105 m/s4105 m/s in a direction making an angle of 6060 with the direction of magnetic field. What will be the pitch of the helix formed by moving particles? (given charge of proton e=1.61019 ce=1.61019 c, mass of proton m=1.671027 kgm=1.671027 kg)
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v1 is responsible for horizontal motion of proton
v2 is responsible for circular motion of proton
∴ mv22 / r = qv2 B
r = mv2 /qB = 1. 76 x 10-27 x 4 x 105 x √3 / 1. 6 x 10-19 x 0.3 x 2 = 0.012 m
Pitch of helix = v1 x T
Where T = 2πr / v2 = 2πr / v sin θ
⇒ Pitch of helix = v cos θ x 2πr / v sin θ
= 2 πr cosθ = 2x 3.14 x 0.012 x cot 60° = 0.044 m
Thanks
v2 is responsible for circular motion of proton
∴ mv22 / r = qv2 B
r = mv2 /qB = 1. 76 x 10-27 x 4 x 105 x √3 / 1. 6 x 10-19 x 0.3 x 2 = 0.012 m
Pitch of helix = v1 x T
Where T = 2πr / v2 = 2πr / v sin θ
⇒ Pitch of helix = v cos θ x 2πr / v sin θ
= 2 πr cosθ = 2x 3.14 x 0.012 x cot 60° = 0.044 m
Thanks
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