A proton with a kinetic energy of $2.0\,eV$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3}\,T$. The angle between the direction of magnetic field and velocity of proton is $60^{\circ}$. 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 $=1.6 \times 10^{-19}\,kg)$
JEE MAIN 2023, Medium
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$B=\frac{\pi}{2} \times 10^{-3}$
$K E=\frac{1}{2} m V^2$
$\Rightarrow V=\sqrt{\frac{2 K E}{m}}$
$=v \cos 60^{\circ} \times \frac{2 \pi m}{e B}$
$=\sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}} \times \cos 60^{\circ} \times \frac{2 \pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \frac{\pi}{2} \times 10^{-3}}$
$=2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5}$
$=4 \times 10^{-1} m =40\,cm$
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