$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

$A$ very long straight wire carries a current $I$. At the instant when a charge $+Q$ at point $P$ has velocity $\vec{V}$,as shown,the force on the charge is

$A$ charged particle is moving in a magnetic field of strength $B$ perpendicular to the direction of the field. If $q$ and $m$ denote the charge and mass of the particle respectively,then the frequency of rotation of the particle is

$A$ proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to the direction of the magnetic field. If both the particles move around the magnetic field in circles of equal radii,the ratio of the momentum of the proton to the alpha particle $\left( \frac{P_p}{P_\alpha} \right)$ is

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?

Uniform magnetic fields of different strengths ($B_1$ and $B_2$),both normal to the plane of the paper,exist as shown in the figure. $A$ charged particle of mass $m$ and charge $q$,at the interface at an instant,moves into the region $2$ with velocity $v$ and returns to the interface. It continues to move into region $1$ and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and $B_2 > B_1$)