$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field 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 $= 1.6 \times 10^{-27} \ kg$)

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j}) \ \text{T}$ has an acceleration $(4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$. The value of $x$ is . . . . . . .

The radius of the circular path of an electron when subjected to a perpendicular magnetic field is:

$A$ proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities,the ratio of the radii $r_p : r_{\alpha}$ of the circular paths described by them will be

$A$ 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: