JEE Advanced 2021 Paper 2 Physics: Paragraph 3

The electric field is parallel to $B$ and to $z$, so it changes only the axial speed and does not affect the circular motion in the $xy$-plane. The perpendicular speed is $v_\perp=v\sin60^\circ=3.464\times10^5\ \mathrm{m\,s^{-1}}$. The magnetic force supplies the centripetal force, $qv_\perp B=mv_\perp^2/r$, so $r=\dfrac{mv_\perp}{qB}=\dfrac{1.67\times10^{-27}\times3.464\times10^5}{1.6\times10^{-19}\times0.20}=1.808\times10^{-2}\ \mathrm{m}$. That is $1.81\ \mathrm{cm}$. The radius stays constant because $v_\perp$ is constant; only the pitch of the helix grows with time as the proton speeds up along $z$. Numerically, $v_\perp B q=1.6\times10^{-19}\times3.464\times10^{5}\times0.2$ gives the force, and dividing the momentum $m v_\perp$ by $qB$ is the quickest route to the radius, which is therefore $1.81\ \mathrm{cm}$ to two decimals.

The time period of the circular motion is independent of speed: $T=\dfrac{2\pi m}{qB}=\dfrac{2\pi\times1.67\times10^{-27}}{1.6\times10^{-19}\times0.20}=3.279\times10^{-7}\ \mathrm{s}$. Along $z$ the proton has initial speed $v_\parallel=v\cos60^\circ=2.0\times10^5\ \mathrm{m\,s^{-1}}$ and constant acceleration $a_z=\dfrac{qE}{m}=\dfrac{1.6\times10^{-19}\times2000}{1.67\times10^{-27}}=1.916\times10^{11}\ \mathrm{m\,s^{-2}}$. The axial displacement in one revolution is $z=v_\parallel T+\tfrac12a_zT^2=6.558\times10^{-2}+1.030\times10^{-2}=7.588\times10^{-2}\ \mathrm{m}$, i.e. $7.59\ \mathrm{cm}$. The first term alone, $6.56$ cm, is the pitch without the electric field, a common wrong answer. Using the larger proton speed after the first revolution would be wrong, since the axial speed grows continuously during the turn. The two contributions are therefore added as uniformly accelerated motion along the axis, independent of the circular part.