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
A particle of charge $q$, mass $m$ enters in a region of magnetic field $B$ with velocity $V_0 \widehat i$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^o$ to its ititial velocity:-
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 particle having the same charge as of electron moves in a circular path of radius $0.5
\,cm$ under the influence of a magnetic field of $0.5\,T.$ If an electric field of $100\,V/m$ makes it to move in a straight path, then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19}\, C$ )
An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall
A charge $Q$ moves parallel to a very long straight wire carrying a current $l$ as shown. The force on the charge is