A proton with a kinetic energy of $2.0\ \text{eV}$ moves into a region of uniform magnetic field of magnitude $\dfrac\pi2\times10^{-3}\ \text{T}$. 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 $=1.6\times10^{-27}$ kg and charge on proton $=1.6\times10^{-19}$ C)
Numerical value type. Enter your answer.
Answer: 40
$v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{2\times3.2\times10^{-19}}{1.6\times10^{-27}}}=2\times10^4\ \text{m s}^{-1}$.
$T=\dfrac{2\pi m}{qB}=\dfrac{2\pi\times10^{-8}}{(\pi/2)\times10^{-3}}=4\times10^{-5}\ \text{s}$.
Pitch $=v\cos60^\circ\,T=10^4\times4\times10^{-5}=0.4\ \text{m}=40\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics