$A$ 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

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.

$A$ proton (mass $m$,charge $q$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space of width $d$. If $\alpha$ is the angle of deviation of the proton from the initial direction of motion (see figure),the value of $\sin \alpha$ will be:

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

If a charged particle moves in a gravity-free space with uniform velocity,then which of the following is not possible? ($\overrightarrow{E} =$ electric field,$\overrightarrow{B} =$ magnetic field)

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?