$A$ proton and an alpha particle moving with equal speeds enter normally into a uniform magnetic field. The ratio of times taken by the proton and the alpha particle to make one complete revolution in the magnetic field is

An electron enters a chamber in which a uniform magnetic field is present as shown in the figure. Ignore gravity. During its motion inside the chamber:

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

$A$ proton (mass $m$,charge $q$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space of width $d$. If $\alpha$ is the angle of deviation of the proton from the initial direction of motion (see figure),the value of $\sin \alpha$ will be:

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

$A$ charged particle is moving in a circular orbit of radius $6 \, cm$ with a uniform speed of $3 \times 10^6 \, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4} \, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is