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 $60^{o}$ . The pitch of the helical path taken by the proton is _________ cm (Take, mass of proton = $1.6 \times 10^{- 27}$ kg and charge on proton = $1.6 \times 10^{- 19} C$ ).

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Detailed Solution

$K . E = 2 e v, B = \frac{π}{2} \times 10^{- 3} π, θ = 60^{o}$

$\frac{1}{2} m v^{2} = 2 e v$

$\frac{1}{2} \times 1.6 \times 10^{- 27} \times v^{2} = 2 \times 1.6 \times 10^{- 19} J$

$v^{2} = 4 \times 10^{8} m / s \Rightarrow v = 2 \times 10^{4} m / s$

Pitch of the helical path $= v_{x} \times T$

$= v \cos θ \times \frac{2 π m}{B q}$

$= \frac{2 π \times 1.6 \times 10^{- 27} \times 2 \times 10^{4} \cos 60^{o}}{\frac{π}{2} \times 1.6 \times 10^{- 19}}$

$= 4 \times 10^{- 1} = 0.4 m$

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