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

$A$ particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform electric and magnetic fields, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

$A$ proton and an alpha particle are separately projected in a region where a uniform magnetic field exists. Their initial velocities are perpendicular to the direction of the magnetic field. If both the particles move around the magnetic field in circles of equal radii,the ratio of the momentum of the proton to the alpha particle $\left( \frac{P_p}{P_\alpha} \right)$ is

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