A particle of mass $m$ and charge $q$ is thrown from origin at $t = 0$ with velocity $2\hat{i}$ + $3\hat{j}$ + $4\hat{k}$ units in a region with uniform magnetic field $\vec B$ = $2\hat{i}$ units. After time $t =\frac{{\pi m}}{{qB}}$ , an electric field is switched on such that particle moves on a straight line with constant speed. $\vec E$ may be
$5\hat k - \,10\hat j \ units$
$-6\hat k - \,9\hat j \ units$
$-6\hat k + \,8\hat j \ units$
$6\hat k + \,8\hat j \ units$
A proton (or charged particle) moving with velocity $v$ is acted upon by electric field $E$ and magnetic field $B$. The proton will move undeflected if
A particle of mass $m,$ charge $Q$ and kinetic energy $K$ enters a transverse uniform magnetic field of induction $B.$ After $3$ $seconds$ the kinetic energy of the particle will be .......$K$
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let $r_p, r_e$ and $r_{He}$ be their respective radii, then
A charged particle of mass $m$ and charge $q$ travels on a circular path of radius $r$ that is perpendicular to a magnetic field $B$. The time taken by the particle to complete one revolution is
A proton and a deutron ( $\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is: