$A$ 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......$^o$

$A$ charge of $4.0 \mu C$ is moving with a velocity of $4.0 \times 10^6 \ m/s$ along the positive $y$-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \ T$. The force acting on the charge is $x \hat{i} \ N$. The value of $x$ is . . . . . . .

An electron revolves around the nucleus in a circular path with angular momentum $\vec{L}$. $A$ uniform magnetic field $\vec{B}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\vec{\tau}$, then

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

At $t = 0$,a positively charged particle of mass $m$ is projected from the origin with velocity $u_0$ at an angle $37^o$ from the $x$-axis as shown in the figure. $A$ constant magnetic field $\vec{B_0} = B_0 \hat{j}$ is present in space. After a time interval $t_0$,the velocity of the particle may be:

$A$ particle of mass $0.6\, g$ and having a charge of $25\, nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4\, m/s$ in a uniform magnetic field. If the particle continues to move with uniform velocity,the value of the magnetic induction is $(g = 10\, m/s^2)$.