An electron having a charge $e$ moves with a velocity $\vec{v}$ in the positive $x$-direction. $A$ magnetic field $\vec{B}$ acts on it in the positive $y$-direction. The force on the electron acts in (where the outward direction is taken as the positive $z$-axis):

An electron is moving at a speed of $3.2 \times 10^7 \ m/s$ in a magnetic field of $12 \times 10^{-4} \ T$ perpendicular to the direction of motion of the electron. The radius of the path of the electron is . . . . . . $cm$. $(e = 1.6 \times 10^{-19} \ C$ and $m_e = 9 \times 10^{-31} \ kg)$

$A$ particle of mass $M$ and positive charge $Q$,moving with a constant velocity $\vec{u}_1 = 4\hat{i} \text{ m/s}$,enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region,the particle emerges on the other side after $10 \text{ ms}$ with a velocity $\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j}) \text{ m/s}$. The correct statement$(s)$ is (are):
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction.
$(C)$ The magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

$A$ proton and a helium nucleus are shot into a magnetic field at right angles to the field with the same kinetic energy. The ratio of their radii is:

$A$ charge of $2.0\,\mu C$ moves with a speed of $3.0 \times 10^6\,m/s$ along the positive $X$-axis. $A$ magnetic field of strength $\vec B = -0.2\,\hat k\,T$ exists in space. What is the magnetic force $(\vec F_m)$ on the charge?

$A$ proton of velocity $(3\hat i + 2\hat j) \, ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k) \, T$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times 10^8 \, C/kg$)