A beam of protons with a velocity 4 x 10 5 m / sec

Question

A beam of protons with a velocity 4 x 105 m / sec enters a uniform magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation )

Navjyot Kalra · Accepted Answer

Hello Student,

Please find the answer to your question

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
Navjot Kalra
askIITians Faculty

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Magnetism> A beam of protons with a velocity 4 x 10 ...

A beam of protons with a velocity 4 x 105 m / sec enters a uniform magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Also find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of rotation )

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