$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

An electron and a proton are moving on straight parallel paths with the same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement$(s)$ is/are true?
$(A)$ They will never come out of the magnetic field region.
$(B)$ They will come out travelling along parallel paths.
$(C)$ They will come out at the same time.
$(D)$ They will come out at different times.

$A$ proton moving with a velocity of $8 \times 10^5 \ m/s$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is $8.3 \ cm$,then the magnitude of the magnetic field is (Charge of proton $= 1.6 \times 10^{-19} \ C$ and mass of the proton $= 1.66 \times 10^{-27} \ kg$) (in $mT$)

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$

$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

$A$ proton and a helium nucleus are shot into a magnetic field at right angles to the field with the same kinetic energy. The ratio of their radii is: