$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. The value of the magnetic induction is $\left(g=10 \,ms^{-2}\right)$.

An electron of mass $m$ and charge $q$ is moving with a speed $v$ in a circular path of radius $r$ perpendicular to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved,what will be the radius of the resulting path?

$A$ 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

The magnetic force acting on a charged particle of charge $-2\, \mu C$ in a magnetic field of $2\, T$ acting in the $y$-direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$,is:

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

$A$ proton, a deuteron, and an $\alpha$-particle enter a region of a uniform magnetic field perpendicular to their velocities with the same kinetic energy. If $r_p, r_d,$ and $r_\alpha$ are the radii of the circular paths of these particles, respectively, then: