Question

A proton moves at a speed $v=2\times {10}^{6}m/s$ in a region of constant magnetic field of magnitude $B=0.05T$. The direction of the proton when it enters this field is $\theta ={30}^{\circ }$ to the field. When you look along the direction of the magnetic field, the path is a circle projected on a plane perpendicular to the magnetic field. How far will the protons move along the direction of $B$ when $2$ projected circles have been completed?

Answer 1

Time of motion of proton is T

$T=\frac{2\pi m_p}{qB\sin {30}^{\circ }}T=\frac{2\times 3.24\times 1.67\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.05\times \frac{1}{2}}T=4.19\times {10}^{-6}sec$

Time taken in two circle $\left(t\right)=2T$

proton cover distance along B is $d=vt$

$d=2\times {10}^6\times 4.19\times {10}^{-6}d=16.78m$