An electron (mass = $9.0 \times 10^{-31} \ kg$ and charge = $1.6 \times 10^{-19} \ C$) is moving in a circular orbit in a magnetic field of $1.0 \times 10^{-4} \ Wb/m^2$. Its period of revolution is:

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

In a crossed field, the magnetic field induction is $2.0 \,T$ and electric field intensity is $20 \times 10^3 \,V/m$. At which velocity will the electron travel in a straight line without being deflected by the electric and magnetic fields?

$A$ proton and an alpha particle moving with energies in the ratio $1: 4$ enter a uniform magnetic field of $3 \ T$ at right angles to the direction of the magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is

$A$ particle of charge $q$ moves with a velocity $\vec{V} = a \hat{i}$ in a magnetic field $\vec{B} = b \hat{j} + c \hat{k}$,where $a$,$b$,and $c$ are constants. The magnitude of the force experienced by the particle is:

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$