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

$A$ charged particle is moving in a uniform magnetic field $(2 \hat{i} + 3 \hat{j}) \text{ T}$. If it has an acceleration of $(\alpha \hat{i} - 4 \hat{j}) \text{ m/s}^2$,then the value of $\alpha$ will be:

Which of the following particles will experience maximum acceleration when projected with the same speed in a transverse magnetic field?

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

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:

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .