Velocity and acceleration vectors of a charged particle moving perpendicular to the direction of a magnetic field at a given instant of time are $\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$ respectively. Then the value of $c$ is

$A$ particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{QB}.$ The time spent by the particle in the field will be:

Two charged particles,having the same kinetic energy,are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of the radii of their circular paths is $6:5$ and their respective masses ratio is $9:4$,then the ratio of their charges will be.

$A$ charged particle is moving in a uniform magnetic field in a circular path. The radius of the circular path is $R$. When the energy of the particle is doubled,the new radius will be:

$A$ 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 $= 1.6 \times 10^{-27} \ kg$)

$A$ positive charge particle having charge $q$ and mass $m$ has velocity $\vec v = v\left( {\frac{{\hat i + \hat k}}{{\sqrt 2 }}} \right)$ in the magnetic field $\vec B = B\hat j$ and electric field $\vec E = E\hat j$ at the origin. Its speed as a function of $y$ is: