$A$ particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$. The radius of the circular path described by it will be:

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

$A$ proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius

$A$ charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be

$A$ particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with a uniform magnetic field $B$ along the $\hat{k}$ direction. The particle will penetrate this region in the $x$-direction up to a distance $d$ equal to:

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: mass of electron $m_e = 9 \times 10^{-31} \ kg$)