Question

A proton of charge $1.6\times {10}^{-19}\text{C}$ and mass $m=1.67\times {10}^{-27}\text{kg}$ is shot with a speed $8\times {10}^6\text{m}/\text{s}$ at an angle of ${30}^{\circ }$with $x$-axis. A uniform magnetic field $B=0.30\text{T}$ exist along the $x$-axis. Find the radius of the helical path followed by the proton ?

• A. $0.241\text{m}$
• B. $0.139\text{m}$
• C. $0.278\text{m}$
• D. $0.482\text{m}$

Answer 1

The correct option is B $0.139\text{m}$

Velocity component of proton along $y$-axis (perpendicular to magnetic field) is responsible for the circular motion,

$v_{\perp }=v\sin {30}^{\circ }=8\times {10}^6\times \left(\frac{1}{2}\right)=4\times {10}^6\text{m}/\text{s}$

Radius of the path, $r=\frac{m{v}_{\perp }}{qB}$

$\Rightarrow r=\frac{1.67\times {10}^{-27}\times 4\times {10}^6}{1.6\times {10}^{-19}\times 0.3}$

$\therefore r=0.139\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.

Why this question ? Tip : Radius of the helical path is calculated using component of velocity which is perpendicular to the magnetic field.