$A$ charged particle moving in a magnetic field experiences a resultant force:

Two charged particles of mass $m$ and charge $q$ each are projected from the origin simultaneously with the same speed $V$ in a transverse magnetic field $B$. If $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of the particles (with respect to the origin) at $t = \frac{\pi m}{qB}$, then the value of $\vec{r}_1 \cdot \vec{r}_2$ at that time is:

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

An electron is moving along the positive $x$-axis. If a uniform magnetic field is applied parallel to the negative $z$-axis,then:
$A.$ The electron will experience a magnetic force along the positive $y$-axis.
$B.$ The electron will experience a magnetic force along the negative $y$-axis.
$C.$ The electron will not experience any force in the magnetic field.
$D.$ The electron will continue to move along the positive $x$-axis.
$E.$ The electron will move along a circular path in the magnetic field.
Choose the correct answer from the options given below:

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

$A$ particle having the same charge as an electron moves in a circular path of radius $0.5 \, cm$ under the influence of a magnetic field of $0.5 \, T$. If an electric field of $100 \, V/m$ makes it move in a straight path,then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19} \, C$):