In a Bainbridge mass spectrograph,a potential difference of $1000 \ V$ is applied between two plates separated by a distance of $1 \ cm$. If the magnetic field is $B = 1 \ T$,what is the velocity of the undeflected positive ions in $m/s$ from the velocity selector?

An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is:

If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$,then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)

In the Thomson experiment for finding $e/m$ for electrons,the beam of electrons is replaced by a beam of muons (particles with the same charge as electrons but with a mass $208$ times that of electrons). The condition for no deflection is satisfied if: