$A$ charge $+Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

$A$ charged particle with specific charge $S$ moves undeflected through a region of space containing mutually perpendicular uniform electric and magnetic fields $E$ and $B$. When the electric field is switched off,the particle will move in a circular path of radius:

$A$ rectangular region of dimensions $(\omega \times l)$ where $\omega \ll l$ has a constant magnetic field $B$ directed into the plane of the paper as shown in the figure. On one side,the region is bounded by a screen. On the other side,positive ions of mass $m$ and charge $q$ are accelerated from rest by a parallel plate capacitor at a constant potential difference $V$ and enter the magnetic field region through a small hole. Which one of the following statements is correct regarding the charge $q$ on the ions that hit the screen?

An electron moves through a uniform magnetic field $\vec{B} = B_0 \hat{i} + 2 B_0 \hat{j} \ T$. At a particular instant of time,the velocity of the electron is $\vec{v} = 3 \hat{i} + 5 \hat{j} \ m/s$. If the magnetic force acting on the electron is $\vec{F} = 5e \hat{k} \ N$,where $e$ is the magnitude of the charge of an electron,then the value of $B_0$ is . . . . . . $T$.

$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is: