A proton projected into a magnetic field of 0.020 T moves in a...

Given:
Magnetic field $B = 0.020 \text{ T}$
Radius $r = 5.0 \text{ cm} = 0.05 \text{ m}$
Pitch $p = 20 \text{ cm} = 0.20 \text{ m}$
Mass of proton $m = 1.6 \times 10^{-27} \text{ kg}$
Charge of proton $q = 1.6 \times 10^{-19} \text{ C}$

The radius of the helical path is given by:
$r = \dfrac{m v_{\perp}}{q B}$

Solving for the perpendicular velocity component $v_{\perp}$:
$v_{\perp} = \dfrac{q B r}{m} = \dfrac{1.6 \times 10^{-19} \times 0.020 \times 0.05}{1.6 \times 10^{-27}} = 1.0 \times 10^5 \text{ m/s}$

The pitch of the helical path is given by:
p = v_{\parallel} T = v_{\parallel} \left(\dfrac{2 \pi m}{q B} \right)

Solving for the parallel velocity component $v_{\parallel}$:
$v_{\parallel} =$\dfrac{p q B}{2 \pi m} = \dfrac{0.20 \times 1.6 \times 10^{-19} \times 0.020}{2 \pi \times 1.6 \times 10^{-27}}

$v_{\parallel} =$\dfrac{0.004}{2 \pi} \times$10^8 =$\dfrac{2}{\pi} \times$10^5 \approx 6.4 \times 10^4 \text{ m/s}$

Thus, the parallel and perpendicular velocity components are $6.4 \times 10^4 \text{ m/s}$ and $1.0 \times 10^5 \text{ m/s}$ respectively.