Particles having positive charges occasionally come with high velocity from the sky towards the earth. On account of the magnetic field of the earth,they would be deflected towards the

Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $\theta$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then,they will meet again for the first time at a point whose distance from the point of projection is

An electron having kinetic energy of $100 eV$ circulates in a path of radius $10 cm$ in a magnetic field. The magnitude of magnetic field $|B|$ is approximately [Mass of electron $= 0.5 MeV/c^2$,where $c$ is the velocity of light].

An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively,in a uniform magnetic field $\vec{B} = B_0 \hat{i}$,each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is