An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become

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

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?

$A$ charged particle in a uniform magnetic field $B = B_0 \hat{k}$ starts moving from the origin with velocity $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$. The trajectory of the particle and the time $t$ at which it reaches $2 \text{ m}$ above the $x-y$ plane are,

$A$ particle having charge of $1 \, C$,mass $1 \, kg$ and speed $1 \, m/s$ enters a uniform magnetic field,having magnetic induction of $1 \, T$,at an angle $\theta = 30^\circ$ between the velocity vector and the magnetic induction. The pitch of its helical path is (in meters):

$A$ $5 \text{ mg}$ particle carrying a charge of $5 \times 10^{-6} \text{ C}$ is moving with a velocity of $(3\hat{i} + 2\hat{k}) \times 10^{-2} \text{ m/s}$ in a region having a magnetic field $\vec{B} = 0.1\hat{k} \text{ Wb/m}^2$. It moves a distance of $\alpha \text{ m}$ along the $\hat{k}$ direction when it completes $5$ revolutions. The value of $\alpha$ is . . . . . . .