A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude $$\dfrac{\pi}{2} \times 10^{-3}$$ 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).

We need to find the pitch of the helical path taken by a proton with kinetic energy 2.0 eV in a magnetic field of magnitude $$\frac{\pi}{2} \times 10^{-3}$$ T, where the angle between the velocity and the magnetic field is 60°.

The kinetic energy relation $$KE = \frac{1}{2}mv^2$$ allows us to solve for the proton’s speed as $$v = \sqrt{\frac{2 \cdot KE}{m}}$$; substituting $$KE = 2.0 \times 1.6 \times 10^{-19}\,\text{J}$$ and $$m = 1.6 \times 10^{-27}\,\text{kg}$$ gives $$v = \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27}}} = \sqrt{4 \times 10^{8}} = 2 \times 10^4\,\text{m/s}$$.

The component of this velocity parallel to the magnetic field is $$v_{\parallel} = v\cos60^\circ = 2 \times 10^4 \times \tfrac12 = 10^4\,\text{m/s}$$.

The proton’s circular motion in the plane perpendicular to the field has period $$T = \frac{2\pi m}{qB}$$, and with $$m = 1.6 \times 10^{-27}\,\text{kg}$$, $$q = 1.6 \times 10^{-19}\,\text{C}$$, and $$B = \frac{\pi}{2} \times 10^{-3}\,\text{T}$$ one finds $$T = \frac{2\pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \frac{\pi}{2} \times 10^{-3}} = \frac{2\pi \times 1.6 \times 10^{-27}}{0.8\pi \times 10^{-22}} = \frac{2 \times 1.6}{0.8} \times 10^{-5} = 4 \times 10^{-5}\,\text{s}$$.

The pitch of the helix follows from multiplying the parallel speed by the period as $$\text{Pitch} = v_{\parallel} \times T = 10^4 \times 4 \times 10^{-5} = 0.4\,\text{m} = 40\,\text{cm}$$, so that the result is $$\boxed{40\ \text{cm}}$$.