An electron of mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ moving with a velocity of $10^6 \ ms^{-1}$ enters a magnetic field normally and describes a circle of radius $10 \ cm$. The intensity of the magnetic field is:

$A$ proton accelerated by a potential difference of $500 \ kV$ flies through a uniform transverse magnetic field of $0.1 \ T$. The field is spread over a region of $1.0 \ cm$ thickness. The angle through which the proton gets deviated from its original direction is (Proton mass $= 1.6 \times 10^{-27} \ kg$ and charge of proton $= 1.6 \times 10^{-19} \ C$) (in $rad$)

$A$ beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20 \ V m^{-1}$ and $0.5 \ T$ at right angles to the direction of motion of the electrons. What is the velocity of the electrons in $m s^{-1}$?

$A$ charged particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ confined between two parallel plates separated by a distance $d$,as shown in the figure. What is the condition for the charged particle not to strike the opposite plate?

$A$ proton of energy $200\, MeV$ enters a magnetic field of $5\, T$. If the direction of the field is from south to north and the motion is upward,the force acting on it will be:

$A$ beam of protons with speed $4 \times 10^{5} \ m/s$ enters a uniform magnetic field of $0.3 \ T$ at an angle of $60^{\circ}$ to the magnetic field. The pitch of the resulting helical path of protons is close to....$cm$
(Mass of the proton $= 1.67 \times 10^{-27} \ kg$, charge of the proton $= 1.6 \times 10^{-19} \ C$)