$A$ positively charged $(+q)$ particle of mass $m$ has kinetic energy $K$ and enters vertically downward into a horizontal magnetic field of induction $\vec{B}$. The acceleration of the particle is:

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

$A$ particle with charge $-Q$ and mass $m$ enters a magnetic field of magnitude $B$, which exists only to the 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,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 $= 9.1 \times 10^{-31} \, kg$)

Which particles will have the minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?