A particle of charge per unit mass $\alpha$ is released from origin with a velocity $\bar{v}=v_0 \vec{i}$ in a uniform magnetic field $\bar{B}=-B_0 \hat{k}$. If the particle passes through $(0, y, 0)$ then $y$ is equal to

Ratio of electric and magnetic field due of moving point charge if its speed is $4.5 \times 10^{5} \;m / s$

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

A $2\, MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5\, tesla$. The force on the proton is

A particle of mass $m = 1.67 \times 10^{-27}\, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $1$ $tesla$ along the direction shown in the figure. The speed of the particle is $10^7\, m/s.$ The magnetic field is directed along the inward normal to the plane of the paper. The particle enters the field at $C$ and leaves at $D.$ Then the angle $\theta$ must be :-.........$^o$

A charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when