A proton with a kinetic energy of $2.0 ~\mathrm{eV}$ moves into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} \mathrm{~T}$. 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 __________ $\mathrm{cm}$. (Take, mass of proton $=1.6 \times 10^{-27} \mathrm{~kg}$ and Charge on proton $=1.6 \times 10^{-19} \mathrm{C}$ ).

<p>Given a proton with a kinetic energy of 2 eV, moving into a region of uniform magnetic field of magnitude $\frac{\pi}{2} \times 10^{-3} T$, and with an angle of $60^{\circ}$ between the direction of the magnetic field and the velocity of the proton, we want to determine the pitch of the helical path taken by the proton.</p> <ol> <li>First, calculate the proton&#39;s speed (v) using the kinetic energy (K.E) formula:</li> </ol> <p>$v = \sqrt{\frac{2 \times KE}{m}}$</p> <ol> <li>Next, find the component of the velocity in the direction of the magnetic field (parallel component):</li> </ol> <p>$v_{\parallel} = v \cos \theta$</p> <p>In this case, θ is given as $60^{\circ}$, so $\cos \theta = \frac{1}{2}$.</p> <ol> <li>The pitch of a charged particle moving in a magnetic field with an angle θ to the direction of the magnetic field is given by the formula:</li> </ol> <p>$p = \frac{2 \pi m v_{\parallel}}{qB}$</p> <ol> <li>Substitute the values for the mass of the proton (m), the kinetic energy (KE), the charge of the proton (q), and the magnetic field (B) into the formula:</li> </ol> <p>$p = \frac{2 \pi \times \sqrt{2mKE} \times \frac{1}{2} \times 2}{qB}$</p> <ol> <li>After substituting the given values, the pitch of the helical path is found to be:</li> </ol> <p>$p = 0.4 \, m = 40 \, cm$</p> <p>In conclusion, the pitch of the helical path taken by the proton in the magnetic field is 40 cm.</p>