Question

A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = 1.6 × 10⁻²⁷ kg

Answer 1

Mass of the proton, m_p = 1.6 × 10⁻²⁷ kg
Magnetic field intensity, B = 0.02 T
Radius of the helical path, r = 5 cm = 5 × 10⁻² m
Pitch of the helical path, p = 20cm = 2 × 10⁻¹ m
We know that for a helical path, the velocity of the proton has two components, $v_{\parallel }\mathrm{and}{v}_{\perp }$.
Now, $\frac{m{v_{\perp }}^2}{r}$ = $q{v}_{\perp }B$
$$\begin{gathered} \Rightarrow r=\frac{m{v}_{\perp }}{qB} \\ \Rightarrow 5\times {10}^{-2}=\frac{1.6\times {10}^{-27}\times {\mathrm{v}}_{\perp }}{1.6\times {10}^{-19}\times 0.02} \\ \Rightarrow {\mathrm{v}}_{\perp }={10}^5\mathrm{m}/\mathrm{s} \end{gathered}$$
$$\begin{gathered} \mathrm{Pitch}=\frac{v_{\parallel }2\mathrm{\pi }r}{v_{\perp }} \\ {v}_{\parallel }=\frac{v_{\perp }P}{2\mathrm{\pi }r} \\ =\frac{{10}^5\times 0.2}{2\times 3.14\times 5\times {10}^{-2}} \\ =0.6369\times {10}^5 \\ =6.4\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$