The radius of the circular path of an electron when subjected to a perpendicular magnetic field is:

Two charges of same magnitude move in two circles of radii $R_1=R$ and $R_2=2R$ in a region of constant uniform magnetic field $B_0$. The work $W_1$ and $W_2$ done by the magnetic field in the two cases respectively,are such that

$A$ magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of $0.75 \;T$. In the same region,a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. $A$ narrow beam of (single species) charged particles all accelerated through $15 \;kV$ enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is $9.0 \times 10^{5} \;V \,m^{-1}$,make a simple guess as to what the beam contains. Why is the answer not unique?

An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ proton enters a magnetic field of flux density $1.5 \,Wb \,m^{-2}$ with a velocity of $2 \times 10^7 \,ms^{-1}$ at an angle of $30^{\circ}$ with the field. The force on the proton will be

$A$ uniform magnetic field $\vec{B} = B_0 \hat{j}$ exists in space. $A$ particle of mass $m$ and charge $q$ is projected towards the negative $x$-axis with speed $v$ from the point $(d, 0, 0)$. The maximum value of $v$ for which the particle does not hit the $y-z$ plane is