$A$ rod $AB$ moves with a uniform velocity $v$ in a uniform magnetic field as shown in the figure.

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths 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 . . . . . . .

$A$ charge $+Q$ is moving upwards vertically. It enters a magnetic field directed to the north. The force on the charge will be towards

An electron is travelling horizontally towards the east. $A$ magnetic field in the vertically downward direction exerts a force on the electron along:

An electric field of $1500\, V/m$ and a magnetic field of $0.40\, Wb/m^2$ act on a moving electron. The minimum uniform speed along a straight line the electron could have is