When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$,it experiences a force $F$. If the potential difference is changed to $V_2$,the force experienced by the electron in the same magnetic field is $2F$. Then,the ratio of potential differences $\frac{V_2}{V_1}$ is:

$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

An electron with kinetic energy $5 \ eV$ enters a region of uniform magnetic field of $3 \ \mu T$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$,so that the electron moves along the same path,is . . . . . $N C^{-1}$.
(Given: mass of electron $= 9 \times 10^{-31} \ kg$,electric charge $= 1.6 \times 10^{-19} \ C$)

Which of the following particles will experience maximum acceleration when projected with the same speed in a transverse magnetic field?