$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The speed of the particle is $v = 10^7 \, m/s$. The magnetic field is directed along the inward normal to the plane of the paper. The particle enters the field at $C$ and leaves at $D$. Then the angle $\theta$ must be :-

$A$ 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 $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field,then the orbit will be

$A$ charged particle moving with a velocity $\vec{v} = v_1 \hat{i} + v_2 \hat{j}$ in a magnetic field $\vec{B}$ experiences a force $\vec{F} = F_1 \hat{i} + F_2 \hat{j}$. Here $v_1, v_2, F_1, F_2$ are all constants. Then $\vec{B}$ can be

Two very long,straight,parallel wires carry steady currents $I$ and $-I$ respectively. The distance between the wires is $d$. At a certain instant of time,a point charge $q$ is at a point equidistant from the two wires,in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is