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?

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 of charge $16 \times 10^{-16} \, C$ moving with velocity $10 \, ms^{-1}$ along the $x$-axis enters a region where a magnetic field of induction $\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \, Vm^{-1}$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $\vec{B}$ is:

$A$ particle with $10^{-11} \, C$ of charge and $10^{-7} \, kg$ mass is moving with a velocity of $10^8 \, m/s$ along the $y$-axis. $A$ uniform static magnetic field $B = 0.5 \, T$ is acting along the $x$-direction. The force on the particle is:

$A$ beam of protons enters a uniform magnetic field of $0.314 \ T$ with a velocity $4 \times 10^5 \ ms^{-1}$ in a direction making an angle $60^{\circ}$ with the direction of the magnetic field. The path of the beam is (mass of proton $= 1.6 \times 10^{-27} \ kg$).

An electron having kinetic energy $T$ is moving in a circular orbit of radius $R$ perpendicular to a uniform magnetic induction $\vec{B}$. If kinetic energy is doubled and magnetic induction is tripled,the radius will become