$A$ charged particle is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. For the magnetic force on it to be maximum,which of the following is correct?

When an electron is accelerated by a potential difference of $V$ volts and enters a magnetic field,the force acting on it is $F$. What will be the force acting on the electron when it is accelerated by a potential difference of $5V$ volts and enters the same magnetic field?

$A$ particle moving with velocity $v$ having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d = \frac{3mv}{5qB}$ at an angle of $53^{\circ}$ to the boundary of the magnetic field. Find the angle $\theta$ shown in the diagram.

$A$ charge $q_0$ moving with velocity $\overrightarrow{v}$ in a magnetic field of induction $\overrightarrow{B}$ experiences a force $\overrightarrow{F}$. The angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $\theta$. The speed of $q_0$ after one second will be

An electron and a proton having same momenta enter perpendicular into a magnetic field then

An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)