$A$ charge of $4.0 \mu C$ is moving with a velocity of $4.0 \times 10^6 \ m/s$ along the positive $y$-axis under a magnetic field $\vec{B}$ of strength $(2 \hat{k}) \ T$. The force acting on the charge is $x \hat{i} \ N$. The value of $x$ is . . . . . . .

$A$ particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent,as shown. The particle after entering the region collides elastically with a rigid wall. The time after which the velocity of the particle becomes anti-parallel to its initial velocity is

$A$ charged particle moves with velocity $\overrightarrow{v}$ in a uniform magnetic field $\overrightarrow{B}$. The magnetic force experienced by the particle is

$A$ particle with charge $-Q$ and mass $m$ enters a magnetic field of magnitude $B$, which exists only to the 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:

$A$ particle of charge $1.6 \ \mu C$ and mass $16 \ \mu g$ is present in a strong magnetic field of $6.28 \ T$. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to its original location for the first time is . . . . . . $s$. $(\pi = 3.14)$

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)