$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ ions having same kinetic energy pass through a region of space filled with a uniform magnetic field $B$ directed perpendicular to the velocity of the ions. The masses of the ions ${H^ + },\,H{e^ + }$ and ${O^{ + + }}$ are respectively in the ratio $1:4:16$. As a result:

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 $.........$

An electron (mass $= 9.1 \times 10^{-31} \, kg$; charge $= -1.6 \times 10^{-19} \, C$) experiences no deflection if subjected to an electric field of $3.2 \times 10^5 \, V/m$ and a magnetic field of $2.0 \times 10^{-3} \, Wb/m^2$. Both the fields are normal to the path of the electron and to each other. If the electric field is removed,then the electron will revolve in an orbit of radius.....$m$:

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then,it travels: