The figure shows three situations when an electron moves with velocity $\vec v$ travels through a uniform magnetic field $\vec B$. In each case, what is the direction of magnetic force on the electron

A proton of energy $8\, eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be.....$eV$

An electron (mass = $9.1 \times {10^{ - 31}}$ $kg$; charge = $1.6 \times {10^{ - 19}}$ $C$) experiences no deflection if subjected to an electric field of $3.2 \times {10^5}$ $V/m$, and a magnetic fields of $2.0 \times {10^{ - 3}} \,Wb/m^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius.......$m$

The magnetic force acting on a charged particle of charge $-2\, \mu  C$ in a magnetic field of $2\, T$ acting in $y$ direction, when the particle velocity is $(2i + 3 j) \times  10^6\,\, m/s$ is

Two long parallel conductors $S_{1}$ and $S_{2}$ are separated by a distance $10 \,cm$ and carrying currents of $4\, A$ and $2 \,A$ respectively. The conductors are placed along $x$-axis in $X - Y$ plane. There is a point $P$ located between the conductors (as shown in figure).

A charge particle of $3 \pi$ coulomb is passing through the point $P$ with velocity

$\overrightarrow{ v }=(2 \hat{ i }+3 \hat{ j }) \,m / s$; where $\hat{i}$ and $\hat{j} \quad$ represents unit vector along $x$ and $y$ axis respectively.

The force acting on the charge particle is $4 \pi \times 10^{-5}(-x \hat{i}+2 \hat{j}) \,N$. The value of $x$ is

At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit

$(i)$ Electrons              $(ii)$ Protons                 $(iii)$ $H{e^{2 + }}$                  $(iv)$ Neutrons

The emission at the instant can be