Question

A proton moves at a speed of $v=2\times {10}^6\text{m/s}$ in a region of constant magnetic field of magnitude $0.05\text{T}$. The direction of the proton with the field is $\theta ={30}^{\circ }$ when it enters. 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 proton moves along the direction of $B$, when two projected circles have bean completed ?

• A. $4.35\text{m}$
• B. $0.209\text{m}$
• C. $2.81\text{m}$
• D. $2.41\text{m}$

Answer 1

The correct option is A $4.35\text{m}$

We know that when $\overrightarrow{v}$ makes an angle $\theta (0<\theta <\pi )$ with the magnetic field $\overrightarrow{B}$, then it follows a helical path.


The distance covered along the direction of $\overrightarrow{B}$, when two circular projection completed;

$d=2\times \text{pitch}$

$d=2v_{||}T=2\left(v\cos \theta \right)T$

$d=2(2\times {10}^6\times \frac{\sqrt{3}}{2})\times \frac{2\pi m}{qB}$

$d=\frac{2\times {10}^6\times \sqrt{3}\times 2\pi \times 1.6\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.05}$

$\therefore d\approx 4.35\text{m}$

Hence, option $\left(a\right)$ is the right choice.