Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

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

Sum

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

#### Solution

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

Concept: Force on a Moving Charge in Uniform Magnetic and Electric Fields
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 12 Magnetic Field
Q 50 | Page 234
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