A particle of charge $q$ and velocity $v$ passes undeflected through a space with non-zero electric field $E$ and magnetic field $B$. The undeflecting conditions will hold if.
signs of both $q$ and $E$ are reversed.
signs of both $q$ and $B$ are reversed.
both $B$ and $E$ are changed in magnitude, but keeping the product of $|B|$ and $|E|$ fixed.
both $B$ and $E$ are doubled in magnitude.
An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5\, R)$ respectively, in a uniform magnetic field ${\rm{\vec B = }}{{\rm{B}}_0}{\rm{\hat i}}$ , each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ?
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)
Two ions having masses in the ratio $1 : 1$ and charges $1 : 2$ are projected into uniform magnetic field perpendicular to the field with speeds in the ratio $2 : 3$. The ratio of the radii of circular paths along which the two particles move is
An $\alpha -$ particle of $1\,MeV$ energy moves on circular path in uniform magnetic field. Then kinetic energy of proton in same magnetic field for circular path of double radius is......$MeV$
A proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,\, MeV,$ the energy acquired by the alpha particle will be......$MeV$