$A$ charged particle $q$ is accelerated by a potential difference of $V$ and enters a region of uniform magnetic field of induction $B$ at right angles to the direction of the field. The charged particle completes a semi-circle of radius $r$ inside the magnetic field. The mass of the charged particle is (all quantities are in $SI$ units):

In the figure shown,a charge $q$ of mass $m$ moving with a velocity $v$ along the $x$-axis enters a region of uniform magnetic field $B$ directed into the page. If the particle is able to enter the region $x > b$,then the velocity $v$ must be greater than:

$A$ charged particle is moving in a uniform magnetic field in a circular path. The radius of the circular path is $R$. When the energy of the particle is doubled,the new radius will be:

$(a)$ $A$ monoenergetic electron beam with electron speed of $5.20 \times 10^{6} \;m s^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} \;T$ normal to the beam velocity. What is the radius of the circle traced by the beam,given $e/m$ for electron equals $1.76 \times 10^{11} \;C \;kg^{-1}$?
$(b)$ Is the formula you employ in $(a)$ valid for calculating the radius of the path of a $20 \;MeV$ electron beam? If not,in what way is it modified?

$A$ beam of ions with velocity $2 \times 10^5 \ m/s$ enters normally into a uniform magnetic field of $4 \times 10^{-2} \ T$. If the specific charge of the ion is $5 \times 10^7 \ C/kg$,then the radius of the circular path described will be $....... \ m$.

$A$ uniform magnetic field $B$ and a uniform electric field $E$ act in a common region. An electron is entering this region of space. The correct arrangement for it to escape undeviated is