A charge moving with velocity $v$ in $X$-direction is subjected to a field of magnetic induction in the negative $X$-direction. As a result, the charge will

A proton and a deutron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field $B$. For motion of proton and deutron on circular path of radius ${R_p}$ and ${R_d}$ respectively, the correct statement is

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed $'v'$ in a uniform magnetic field $B$ going into the plane of the paper (See figure). If charge densities ${\sigma _1}$ and ${\sigma _2}$ are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects)

A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be

A positively charged particle enters in a region of uniform. Transverse magnetic field as shown in figure find net deviation in path of the particle.