If a proton enters perpendicularly into a magnetic field with velocity $v$,the time period of revolution is $T$. If the proton enters with velocity $2v$,what will be the time period?

The magnetic field is uniform for $y > 0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y < 0$. $A$ proton,denoted by a filled circle,leaves $y = 0$ in the $-y$-direction with some speed,as shown below. Which of the following best denotes the trajectory of the proton?

$A$ proton and a deuteron are both accelerated through the same potential difference and enter a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$,assuming the neutron and proton masses are nearly equal,the radius of the proton's path will be

$A$ proton and a deuteron,both having the same kinetic energy,enter perpendicularly into a uniform magnetic field $B$. For the motion of the proton and deuteron on circular paths of radii ${R_p}$ and ${R_d}$ respectively,the correct statement is:

When a positively charged particle enters a uniform magnetic field with uniform velocity, its trajectory can be:
$(1)$ a straight line
$(2)$ a circle
$(3)$ a helix

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)