If cathode rays are projected at right angles to a magnetic field,their trajectory is

An electron of mass $m$ and charge $q$ is moving with a speed $v$ in a circular path of radius $r$ perpendicular to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved,what will be the radius of the resulting path?

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

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 $= 1.6 \times 10^{-19} \, C$)

$A$ deuteron of kinetic energy $50 \, keV$ is describing a circular orbit of radius $0.5 \, m$ in a plane perpendicular to a magnetic field $\vec{B}$. The kinetic energy of a proton that describes a circular orbit of radius $0.5 \, m$ in the same plane with the same $\vec{B}$ is ........ $keV$.

$A$ charge $q_0$ moving with velocity $\overrightarrow{v}$ in a magnetic field of induction $\overrightarrow{B}$ experiences a force $\overrightarrow{F}$. The angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $\theta$. The speed of $q_0$ after one second will be