If the magnetic field is parallel to the positive $y-$axis and the charged particle is moving along the positive $x-$axis (Figure), which way would the Lorentz force be for

$(a)$ an electron (negative charge),

$(b)$ a proton (positive charge).

An electron is moving with a speed of ${10^8}\,m/\sec $ perpendicular to a uniform magnetic field of intensity $B$. Suddenly intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$

A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?

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 $ = 1.7 \times {10^{ – 27}}\,kg$ and charge $ = 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 charged particle enters a magnetic field $H$ with its initial velocity making an angle of $45^\circ $ with $H$. The path of the particle will be