A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude $\frac{π}{2} \times 10^{- 3} T .$ The angle between the direction of magnetic field and velocity of proton is $60^{0}$ . The pitch of the helical path taken by the proton is ____cm (Take, masss of proton= $1.6 \times 10^{- 27} k g$ and Charge on proton = $1.6 \times 10^{- 19} C$ .

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Detailed Solution

$K . E = 2 ev, B = \frac{π}{2} \times 10^{- 3} π, θ = 60^{0} \frac{1}{2} {mv}^{2} = 2 ev \frac{1}{2} \times 1.6 \times 10^{- 27} \times v^{2} = 2 \times 1.6 \times 10^{- 19} J v^{2} = 4 \times 10^{8} m / s \Rightarrow v = 2 \times 10^{4} m / s Pitch of the helical path = v_{x} \times T = vcosθ \times \frac{2 πm}{Bq} = \frac{2 π \times 1.6 \times 10^{- 27} \times 2 \times 10^{4} \cos 60^{0}}{\frac{π}{2} \times 10^{- 3} X 1.6 \times 10^{- 19}} = 4 \times 10^{- 1} = 0.4 m$

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