Question

Two identical charged particles enter a uniform magnetic field with same speed but at angles ${30}^o$ and ${60}^o$ with field. Let a, b and c be the ratio of their time periods, radii and pitches of the helical paths then:

• A. $abc=1$
• B. $abc>1$
• C. $abc<1$
• D. $a=bc$

Answer 1

Answer: (A), (D)

The correct options are
A $abc=1$
D $a=bc$

Component of velocity ${\perp }^r$ to the $\overrightarrow{B}$ will make path of particle circular and component of velocity parallel to magnetic field will make its path helical.

as time period is given by

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

$T_1=\frac{2\pi m}{qB}$ $T_2=\frac{2\pi m}{qB}$

$\therefore a=\frac{T_1}{T_2}=1$

radius is given by

$r=\frac{m\vartheta }{qB}$

$\therefore r_1=\frac{m\vartheta sin{30}^{\circ }}{qB}=\frac{m\vartheta }{2qB}$

and $r_2=\frac{m\vartheta sin{60}^{\circ }}{qB}=\frac{\sqrt{3}m\vartheta }{2qB}$

$\therefore b=\frac{r_1}{r_2}=\frac{1}{\sqrt{3}}$

Pitch is given by $=T\ast {\vartheta }_{parallel}$

$P_1=\frac{2\pi m}{qB}\vartheta cos{30}^{\circ }=\frac{2\pi m\vartheta }{qB}\ast \frac{\sqrt{3}}{2}$

$P_2=\frac{2\pi m}{qB}\vartheta cos{60}^{\circ }=\frac{2\pi m\vartheta }{qB}\ast \frac{1}{2}$

$\therefore c=\frac{P_1}{P_2}=\frac{\sqrt{3}}{1}$

$\therefore abc=1\ast \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{1}=1$