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A Proton Projected in a Magnetic Field of 0.020 T Travels Along a Helical Path of Radius 5.0 Cm and Pitch - Physics

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प्रश्न

A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10−27 kg

योग

उत्तर

Mass of the proton, mp = 1.6 × 10−27 kg
Magnetic field intensity, B = 0.02 T
Radius of the helical path, r = 5 cm = 5 × 10−2 m
Pitch of the helical path, p = 20cm = 2 × 10−1 m
We know that for a helical path, the velocity of the proton has two components, 
`v_"||"and v ⊥.`
Now , `(mv ⊥^2)/r = qv ⊥B`
⇒ r = `(mv ⊥)/(r) = qv ⊥ B`
⇒ `5xx10^-2 = (1.6xx10^-27xxv_1)`
⇒ `v ⊥ = 10^5 m //s`
Pitch = `v_"||" = (v_1P)/(2pir)`
`=(10^5xx0.2)/(2xx3.14xx5xx10^-2)`
= 0.6369 × 10^5
= 6.4 × 10^4 m/s

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Force on a Moving Charge in Uniform Magnetic and Electric Fields
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अध्याय 12: Magnetic Field - Exercises [पृष्ठ २३४]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 12 Magnetic Field
Exercises | Q 50 | पृष्ठ २३४

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