The direction of the magnetic force on the electron as shown in the diagram is along:

In an experiment,electrons are accelerated from rest by applying a voltage of $500 \, V$. Calculate the radius of the path if a magnetic field of $100 \, mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19} \, C$,Mass of the electron $= 9.1 \times 10^{-31} \, kg$]

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

$A$ charged particle moving along a straight line path enters a uniform magnetic field of $4 \ mT$ at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7 \ C \ kg^{-1}$,the angular velocity of the particle in the magnetic field is

$A$ particle having charge $q$ enters a region of uniform magnetic field $\vec{B}$ (directed inwards) and is deflected a distance $x$ after travelling a distance $y$. The magnitude of the momentum of the particle is:

$A$ particle of mass $m$ and charge $q$ is thrown from the origin at $t = 0$ with velocity $\vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ units in a region with a uniform magnetic field $\vec{B} = 2\hat{i}$ units. After time $t = \frac{\pi m}{qB}$,an electric field $\vec{E}$ is switched on such that the particle moves in a straight line with constant speed. $\vec{E}$ may be: