$A$ charged particle is projected with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$. For the magnetic force on it to be maximum,which of the following is correct?

$A$ proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2 \times 10^5 \text{ ms}^{-1}$. When the electric field is switched off,the proton moves along a circular path of radius $2 \text{ cm}$. The magnitude of the electric field is $x \times 10^4 \text{ N/C}$. The value of $x$ is . . . . . . . (Take the mass of the proton $= 1.6 \times 10^{-27} \text{ kg}$ and charge $e = 1.6 \times 10^{-19} \text{ C}$)

Explain: Velocity selector.

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: If oxygen ion $(O^{-2})$ and hydrogen ion $(H^{+})$ enter normal to the magnetic field with equal momentum,then the path of $O^{-2}$ ion has a smaller curvature than that of $H^{+}$.
Reason $R$: $A$ proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statement,choose the correct answer from the options given below.

$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the energy of the particle becomes three times the original,the new radius will be

$A$ charge '$q$' moves with a velocity $2 \ m/s$ along the $x$-axis in a uniform magnetic field $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$. The charge will experience a force: