Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B$ = $B_0\hat{k}$
They have equal $z$-components of momenta
They must have equal charges
They necessarily represent a particleantiparticle pair
The charge to mass ratio satisfy:${\left( {\frac{e}{m}} \right)_1} + {\left( {\frac{e}{m}} \right)_2} = 0$
A particle of mass $m$ and charge $q$ is thrown from origin at $t = 0$ with velocity $2\hat{i}$ + $3\hat{j}$ + $4\hat{k}$ units in a region with uniform magnetic field $\vec B$ = $2\hat{i}$ units. After time $t =\frac{{\pi m}}{{qB}}$ , an electric field is switched on such that particle moves on a straight line with constant speed. $\vec E$ may be
$OABC$ is a current carrying square loop an electron is projected from the centre of loop along its diagonal $AC$ as shown. Unit vector in the direction of initial acceleration will be
At $t = 0$ a charge $q$ is at the origin and moving in the $y-$ direction with velocity $\overrightarrow v = v\,\hat j .$ The charge moves in a magnetic field that is for $y > 0$ out of page and given by $B_1 \hat z$ and for $y < 0$ into the page and given $-B_2 \hat z .$ The charge's subsequent trajectory is shown in the sketch. From this information, we can deduce that
An electron and a proton enter region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius ${r_e}$ and ${r_p}$ respectively. Then
A small block of mass $m$, having charge $q$ is placed on frictionless inclined plane making an angle $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but perpendicular to the length of spring. If $m$ is slightly pulled on the inclined in downward direction and released, the time period of oscillation will be (assume that the block does not leave contact with the plane)