Question

A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of $200V{m}^{-1}$ makes the path straight . Find the charge / mass ratio of the particle.

Answer 1

Let charge on particle is $q$ and mass of particle is $m$. Moving with velocity $v$, perpendicular to magnetic field $\left(B\right)$ now. Radius of circle related as $\frac{m{V}^2}{r}=qVB$

$V=\frac{q}{m}Br$

If electrostatics force balance electromagnetic force however $q\overrightarrow{E}=qVB$

$\Rightarrow \overrightarrow{E}=VB=\frac{q}{m}Br\left(B\right)$

$\Rightarrow \overrightarrow{E}=\left(\frac{q}{m}\right)B^{2}r$

$\frac{q}{m}=\frac{\overrightarrow{E}}{B^{2}r}=\frac{200}{\left(0.4{)}^2\right(\frac{1\times {10}^{-2}}{2})}=\frac{400}{0.16\times {10}^{-2}}$

$\frac{q}{m}=250\times {10}^3=250k$