Question

A charged particle enters a magnetic field at right angles to the field. The field exists for a length equal to 1.5 times the radius of circular path of particle. The particle will be deviated from its path by :

• A. ${90}^o$
• B. $si{n}^{-1}\left(\frac{2}{3}\right)$
• C. ${30}^o$
• D. ${180}^o$

Answer 1

The correct option is D ${180}^o$

Let magnetic field strength be $B$

Charge $q$

Mass $m$ and

Radius of the circular path be $r$

$\overrightarrow{B=B\hat{k}}$

$\overrightarrow{v}=v\cos \theta \hat{i}+v\sin \theta \hat{j}\overrightarrow{F}=(v\cos \theta \hat{i}+v\sin \theta \hat{j})\times Bq\hat{k}\Rightarrow \overrightarrow{F}=Bqv(\cos \theta \hat{i}+\sin \theta \hat{j})\theta =\omega t{a}_x=\frac{Bqv}{m}\sin \omega t\frac{d^{2}x}{d{t}^2}=\frac{Bqv}{m}\sin \omega t$

Integrating

$\Rightarrow x=\frac{Bqv}{m{\omega }^2}\sin (\omega t+\theta )$

So magnitude of maximum horizontal displacement suffered by the particle in the field will be

$x_{max}=\frac{Bqv}{m{\omega }^2}$

Again,

$x_{max}=\frac{Bqv}{m{\omega }^2}Bqv=m{\omega }^{2}r\omega =\frac{Bq}{m}{x}_{max}=\frac{v}{\omega }$

$v=r\omega \Rightarrow x_{max}=r$

This means for a particle to reach maximum displacement the field must have a minimum length of $r$

since length of field is $1.5r$ the particle cannot come out of field horizontally so to come out of it must reverse its direction and thus displacement.

Thus the particle will deflect by $180°$