A proton moves at 5.0 × 10⁶ m/s in a uniform field of 0.30 T…

Step 1. The two parts. $v_\perp = v\sin 20^\circ = 5.0 \times 10^{6} \times 0.342 = 1.71 \times 10^{6}\ \text{m/s}$. $v_\parallel = v\cos 20^\circ = 5.0 \times 10^{6} \times 0.940 = 4.70 \times 10^{6}\ \text{m/s}$.

Step 2. The radius. The top of $r = \dfrac{mv_\perp}{eB}$ is $mv_\perp = 1.67 \times 10^{-27}\ \text{kg} \times 1.71 \times 10^{6}\ \text{m/s} = 2.86 \times 10^{-21}\ \text{kg m/s}$. The bottom is $eB = 1.6 \times 10^{-19}\ \text{C} \times 0.30\ \text{T} = 4.8 \times 10^{-20}\ \text{C T}$. So $r = \dfrac{2.86 \times 10^{-21}\ \text{kg m/s}}{4.8 \times 10^{-20}\ \text{C T}} = 5.95 \times 10^{-2}\ \text{m}$, that is $5.95$ cm.

Step 3. The period. $T = \dfrac{2\pi m}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27}\ \text{kg}}{4.8 \times 10^{-20}\ \text{C T}} = 2.19 \times 10^{-7}\ \text{s}$.

Step 4. The pitch. $p = v_\parallel T = 4.70 \times 10^{6}\ \text{m/s} \times 2.19 \times 10^{-7}\ \text{s} = 1.03\ \text{m}$.

Step 5. The turns. $N = \dfrac{5.0\ \text{m}}{1.03\ \text{m}} = 4.87$.