$A$ particle having charge $q$ enters a region of uniform magnetic field $\vec{B}$ (directed inwards) and is deflected a distance $x$ after travelling a distance $y$. The magnitude of the momentum of the particle is:

An electron (mass $= 9.1 \times 10^{-31} \, kg$; charge $= -1.6 \times 10^{-19} \, C$) experiences no deflection if subjected to an electric field of $3.2 \times 10^5 \, V/m$ and a magnetic field of $2.0 \times 10^{-3} \, Wb/m^2$. Both the fields are normal to the path of the electron and to each other. If the electric field is removed,then the electron will revolve in an orbit of radius.....$m$:

$A$ strong magnetic field is applied on a stationary electron. Then,the electron

$A$ charge '$q$' moves with a velocity $2 \ m/s$ along the $x$-axis in a uniform magnetic field $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$. The charge will experience a force:

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?

$A$ 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: