When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are
$\frac{{m{a_0}}}{e}$ west , $\frac{{m{a_0}}}{{e{V_0}}}$ up
$\;\frac{{m{a_0}}}{e}$ west , $\frac{{2m{a_0}}}{{e{V_0}}}$ down
$\frac{{m{a_0}}}{e}$ east , $\frac{{3m{a_0}}}{{e{V_0}}}$ up
$\frac{{m{a_0}}}{e}$ east, $\frac{{3m{a_0}}}{{e{V_0}}}$ down
An electron is travelling horizontally towards east. A magnetic field in vertically downward direction exerts a force on the electron along
A charge moving with velocity $v$ in $X$-direction is subjected to a field of magnetic induction in the negative $X$-direction. As a result, the charge will
A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?
A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) T$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) m / s ^{2}$, then the value of $\alpha$ will be.
An electron is projected with velocity $\vec v$ in a uniform magnetic field $\vec B$ . The angle $\theta$ between $\vec v$ and $\vec B$ lines between $0^o$ and $\frac{\pi}{2}$ . It velocity $\vec v$ vector returns to its initial value in time interval of