$A$ 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?

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 $= 9.1 \times 10^{-31} \, kg$)

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

An electron emitted by a heated cathode and accelerated through a potential difference of $2.0 \; kV$,enters a region with a uniform magnetic field of $0.15 \; T$. Determine the trajectory of the electron if the field:
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^{\circ}$ with the initial velocity.

$A$ 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 $mT$)

$A$ particle of mass $m$ and charge $q$ is thrown from the origin at $t = 0$ with velocity $\vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ units in a region with a uniform magnetic field $\vec{B} = 2\hat{i}$ units. After time $t = \frac{\pi m}{qB}$,an electric field $\vec{E}$ is switched on such that the particle moves in a straight line with constant speed. $\vec{E}$ may be: