$A$ uniform magnetic field acts at right angles to the direction of motion of electrons. As a result,the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled,then the radius of the circular path will be.....$cm$.

$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

The direction of the magnetic force on the electron as shown in the diagram is along:

An electron having charge $1.6 \times 10^{-19} \ C$ and mass $9 \times 10^{-31} \ kg$ is moving with $4 \times 10^6 \ m \ s^{-1}$ speed in a magnetic field of $2 \times 10^{-1} \ T$ in a circular orbit. The force acting on the electron and the radius of the circular orbit is $.........$