$A$ mixed beam of $He^+$ and $O^{2+}$ ions (mass of $He^+ = 4 \ amu$ and that of $O^{2+} = 16 \ amu$) passes through a region of a constant perpendicular magnetic field. If the kinetic energy of all the ions is the same,then:

An electron $(e^-)$ is moving parallel to a long current-carrying wire as shown in the figure. Calculate the magnetic force acting on the electron. (Given: $I = 10 \ A$,$v = 10^5 \ m/s$,$r = 4 \ cm = 0.04 \ m$)

$A$ charged particle moves with some initial velocity along the direction of an external magnetic field $B$. Now, if we apply a uniform electric field $E$ perpendicular to the magnetic field, then the trajectory of the charged particle will be

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

The figure shows a region of length $l$ with a uniform magnetic field of $0.3 \, T$ in it. $A$ proton enters the region with a velocity of $4 \times 10^{5} \, m/s$,making an angle of $60^{\circ}$ with the field. If the proton completes $10$ revolutions by the time it crosses the region shown,$l$ is close to....... $m$ (mass of proton $= 1.67 \times 10^{-27} \, kg$,charge of the proton $= 1.6 \times 10^{-19} \, C$)

The motion of a charged particle can be used to distinguish between a magnetic field and an electric field in a certain region by firing the charge: