The magnetic force acting on a charged particle of charge $2\,\mu C$ in a magnetic field of $2\, T$ acting in the $y-$ direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$ is:

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

An electron moves at right angles to a magnetic field of $1.5 \times 10^{-2} \, T$ with a speed of $6 \times 10^7 \, m/s$. If the specific charge of the electron is $1.7 \times 10^{11} \, C/kg$,the radius of the circular path will be ...... $cm$.

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

If a particle of charge $10^{-12} \, C$ moving along the $\hat{x}$ direction with a velocity $10^5 \, m/s$ experiences a force of $10^{-10} \, N$ in the $\hat{y}$ direction due to a magnetic field,then the minimum magnetic field is:

An electron emitted by a heated cathode and accelerated through a potential difference of $2.0 \; kV$,enters a region with a uniform magnetic field of $0.15 \; T$. Determine the trajectory of the electron if the field:
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^{\circ}$ with the initial velocity.