The acceleration of an electron at a moment in a magnetic field $\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ is $\vec{a} = x\hat{i} - 2\hat{j} + \hat{k}$. The value of $x$ is

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

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

$A$ uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5 \ Wb/m^2$. If a proton having mass $m = 1.7 \times 10^{-27} \ kg$ and charge $q = 1.6 \times 10^{-19} \ C$ moves in this field vertically downwards with energy $5 \ MeV$,then the force acting on it will be

$H^+$,$He^+$,and $O^{++}$ are projected in a uniform transverse magnetic field with equal accelerating potential. If their masses are $1 \, a.m.u.$,$4 \, a.m.u.$,and $16 \, a.m.u.$ respectively,find the ratio of their radii.

An electron (mass $= 9.1 \times 10^{-31} \, kg$; charge $= -1.6 \times 10^{-19} \, C$) experiences no deflection if subjected to an electric field of $3.2 \times 10^5 \, V/m$ and a magnetic field of $2.0 \times 10^{-3} \, Wb/m^2$. Both the fields are normal to the path of the electron and to each other. If the electric field is removed,then the electron will revolve in an orbit of radius.....$m$: