$A$ uniform electric field and a uniform magnetic field are produced,pointing in the same direction. An electron is projected with its velocity pointing in the same direction.

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 projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

An electron is moving with a speed of $10^8 \ m/s$ perpendicular to a uniform magnetic field of intensity $B$. Suddenly,the intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$:

$A$ narrow electron beam passes undeviated through an electric field $E = 3 \times 10^4 \ V/m$ and an overlapping magnetic field $B = 2 \times 10^{-3} \ Wb/m^2$. If the electric field and magnetic field are mutually perpendicular,the speed of the electrons is:

$A$ charge $q$ is moving in a magnetic field. Then,the magnetic force does not depend upon: