$A$ particle of mass $M$ and charge $q$ moves with a constant velocity $V$ in the positive $x$-direction. $A$ constant magnetic field $B$ exists in the negative $z$-direction between $x = a$ and $x = b$. Find the minimum value of $V$ such that the particle enters the region $x > b$.

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.
Reason $(R) :$ The magnetic field in that region is along the direction of velocity of the electron.
In the light of the above statements,choose the correct answer from the options given below.

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then,it travels:

If a charged particle moves in a gravity-free space with uniform velocity,then which of the following is not possible? ($\overrightarrow{E} =$ electric field,$\overrightarrow{B} =$ magnetic field)

$A$ particle is projected with a velocity of $10 \ m/s$ along the $y-$axis from the point $(2, 3)$. $A$ uniform magnetic field of $(3\hat{i} + 4\hat{j}) \ T$ exists in the space. What is its speed when the particle passes through the $y-$axis for the third time? (Neglect gravity)