$A$ 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:

$A$ 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 of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The angle of incidence is $45^{\circ}$ and the angle of emergence is also $45^{\circ}$. The time spent by the particle in the magnetic field is $...... \, ns$.

An electron,a proton,a deuteron,and an alpha particle,each having the same speed,are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radii of the circular orbits of these particles are respectively $R_e, R_p, R_d$,and $R_\alpha$. It follows that:

The magnetic field vector of an electromagnetic wave is given by $\vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t)$,where $\hat{i}$ and $\hat{j}$ represent unit vectors along the $x$ and $y$-axes,respectively. At $t = 0 \, s$,two electric charges $q_1 = 4\pi \, C$ and $q_2 = 2\pi \, C$ are located at $(0, 0, \pi/k)$ and $(0, 0, 3\pi/k)$,respectively. Both charges have the same velocity $\vec{v} = 0.5c\hat{i}$,where $c$ is the speed of light. The ratio of the magnetic force acting on charge $q_1$ to that on $q_2$ is:

Two particles $A$ and $B$ have equal charges but different masses $M_{A}$ and $M_{B}$. After being accelerated through the same potential difference,they enter a region of uniform magnetic field and describe paths of radii $R_{A}$ and $R_{B}$ respectively. Then $M_{A} : M_{B}$ is