$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

$A$ monoenergetic $(18 \;keV)$ electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \;G$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \;cm$ $(m_{e} = 9.11 \times 10^{-31} \;kg)$.

An electron and a proton enter a 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:

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $A$ lies in the same plane at a distance $2R$ from the centre. Point $A$ records the magnetic field produced by the $\alpha$ particle. If the minimum time interval between two successive times at which $A$ records zero magnetic field is $t$,find the angular speed $\omega$ in terms of $t$.

An electron and a proton have equal kinetic energies. They enter a magnetic field perpendicularly. Then:

The distance between two plates is $1 \ cm$ and the potential difference is $1000 \ V$. $A$ magnetic field $B = 1 \ T$ is applied. If an electron passes through without any deflection,what will be its velocity?