Question

A proton projected in a magnetic field of $0.020T$ travels along a helical path of radius $5.0cm$ and pitch $20cm$. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton = $1.6\times {10}^{-27}kg$.

Answer 1

$\overrightarrow{B}=0.020T$ $M_p=1.6\times {10}^{-27}kg$

Pitch $=20cm=2\times {10}^{-1}m$

Radius $=5cm=5\times {10}^{-2}m$

We know for a helical path, the velocity of the proton has got two components $v_{\perp }$ and $v_H$

Now, $r=\frac{m{v}_{\perp }}{qB}\Rightarrow 5\times {10}^{-2}=\frac{1.6\times {10}^{-27}\times v_{\perp }}{1.6\times {10}^{-19}\times 2\times {10}^{-2}}$

$\Rightarrow v_{\perp }=\frac{5\times {10}^{-2}\times 1.6\times {10}^{-19}\times 2\times {10}^{-2}}{1.6\times {10}^{-27}}=1\times {10}^{5}m/s$

However, $v_H$ remains constant

$T=\frac{2\pi m}{qB}$

Pitch $=v_H\times T$ or $v_H=\frac{Pitch}{T}$

$v_H=\frac{2\times {10}^{-1}}{2\times 3.14\times 1.6\times {10}^{-27}}\times 1.6\times {10}^{-19}\times 2\times {10}^{-2}=0.6369\times {10}^5\approx 6.4\times {10}^{4}m/s$