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

$A$ $1 \mu\text{C}$ charge is moving with velocity $\vec{v} = (\hat{i} - 2\hat{j} + 3\hat{k}) \text{ m/s}$ in a region of magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 5\hat{k}) \text{ T}$. The magnitude of the force acting on it is $\sqrt{\alpha} \times 10^{-6} \text{ N}$. The value of $\alpha$ is . . . . . . .

An electron is moving along the positive $x$-axis. If a uniform magnetic field is applied parallel to the negative $z$-axis,then:
$A.$ The electron will experience a magnetic force along the positive $y$-axis.
$B.$ The electron will experience a magnetic force along the negative $y$-axis.
$C.$ The electron will not experience any force in the magnetic field.
$D.$ The electron will continue to move along the positive $x$-axis.
$E.$ The electron will move along a circular path in the magnetic field.
Choose the correct answer from the options given below:

$A$ particle moving with velocity $v$ having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d = \frac{3mv}{5qB}$ at an angle of $53^{\circ}$ to the boundary of the magnetic field. Find the angle $\theta$ shown in the diagram.

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

Which of the following statements is true?