An electron beam passes through a magnetic field of $2 \times 10^{-3}\,Wb/m^2$ and an electric field of $1.0 \times 10^4\,V/m$ both acting simultaneously. The path of electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively

An electron, moving along the $x-$ axis with an initial energy of $100\, eV$, enters a region of magnetic field $\vec B = (1.5\times10^{-3}T)\hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2\, cm$. The electron is detected at the point $Q$ on a screen placed $8\, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :……$cm$ (electron's charge $= 1.6\times10^{-19}\, C$, mass of electron $= 9.1\times10^{-31}\, kg$)

Two parallel wires in the plane of the paper are distance $X _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have direction opposite to each other, the radius of curvature of the path is $R_2$.

If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.

Two protons $A$ and $B$ move parallel to the $x$-axis in opposite directions with equal speeds $v$. At the instant shown, the ratio of magnetic force and electric force acting on the proton $A$ is ( $c=$ speed of light in vacuum)

If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity $v$ and an electron is projected along the lines of force, what will happen to proton and electron