$A$ charge $Q$ moves parallel to a very long straight wire carrying a current $I$ as shown. The force on the charge is

$A$ charged particle moving in a magnetic field $B$ has velocity components both along $B$ and perpendicular to $B$. The path of the charged particle will be:

$A$ charged particle is released from rest in a region of steady uniform electric and magnetic fields which are parallel to each other. The particle will move in a:

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ($m$ - mass of the particle,$v$ - velocity of the particle,$q$ - charge of the particle,$B$ - magnetic field).

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly into a region of uniform magnetic field $B$ and executes a circular arc of radius $R$. Then,$\frac{q}{m}$ equals: