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 $r$:

$A$ particle with charge-to-mass ratio $\frac{q}{m} = \alpha$ is shot with a speed $v$ towards a wall at a distance $d$,moving perpendicular to the wall. The minimum value of the magnetic field $\vec{B}$ that exists in this region,perpendicular to the velocity of the particle,such that the particle does not hit the wall is:

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

$A$ proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1 \, MeV,$ the energy acquired by the alpha particle will be......$MeV$

$A$ charge $Q$ is moving through a distance $d\vec{l}$ in a magnetic field $\vec{B}$. Find the value of the work done by the magnetic field.

$A$ beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20 \ V m^{-1}$ and $0.5 \ T$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons in $m s^{-1}$?