$A$ magnetic field $\overrightarrow{B} = B_0 \hat{j}$ exists in the region $a < x < 2a$ and $\overrightarrow{B} = -B_0 \hat{j}$ in the region $2a < x < 3a$,where $B_0$ is a positive constant. $A$ positive point charge moving with a velocity $\overrightarrow{v} = v_0 \hat{i}$,where $v_0$ is a positive constant,enters the magnetic field at $x = a$. The trajectory of the charge in this region can be like,

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

An electron enters with a velocity $\vec{v} = v_0 \hat{i}$ into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in a plane parallel to the $xy$-plane. Suggest a configuration of fields $\vec{E}$ and $\vec{B}$ that can lead to this.

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. The time periods of rotation will be:

An electron,a proton,a deuteron,and an $\alpha -$ particle enter a uniform magnetic field normally with the same velocity. Which particle will have the maximum rotational frequency?

$A$ proton and a deuteron $(q=+e, m=2.0 \ u)$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$,moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of the deuteron path to the radius $r_p$ of the proton path is: