If an electron and a proton having same momenta enter perpendicular to a magnetic field,then

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is $\sigma$. Electric intensity in the space between the plates is $E$. $A$ uniform magnetic field $B$ also exists in that space perpendicular to the direction of $E$. The electron moves perpendicular to both $\vec{E}$ and $\vec{B}$ without any change in direction. The time taken by the electron to travel a distance $\ell$ in the space is

$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

$A$ charged particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,mT$ at a right angle to the direction of the field. The charged particle completes a semicircle of radius $3\,cm$ inside the magnetic field. The mass of the charged particle is $........\times 10^{-18}\,kg$.

An electron with kinetic energy $5 \ eV$ enters a region of uniform magnetic field of $3 \ \mu T$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$,so that the electron moves along the same path,is . . . . . $N C^{-1}$.
(Given: mass of electron $= 9 \times 10^{-31} \ kg$,electric charge $= 1.6 \times 10^{-19} \ C$)

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.