A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude . The angle between the direction of magnetic field and velocity of proton is . [2023]

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Solution

Q.
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 $60 °$ . 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 $= 1.6 \times 10^{- 19} C$ .) [2023]

Ans.
(40)

$B = \frac{π}{2} \times 10^{- 3}$

$K.E. = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 K.E.}{m}}$

$Pitch = v \cos 60 ° \times time period of one rotation$

$= v \cos 60 ° \times \frac{2 π m}{e B}$

$= \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{- 19}}{1.6 \times 10^{- 27}}} \times \cos 60 ° \times \frac{2 π \times 1.6 \times 10^{- 27}}{1.6 \times 10^{- 19} \times \frac{π}{2} \times 10^{- 3}}$

$= 2 \times 10^{4} \times \frac{1}{2} \times 4 \times 10^{- 5}$

$= 4 \times 10^{- 1} m = 40 cm$

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