A proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha - $particle (mass = $4m$ and charge = $+ 2e),$ so that it can revolve in the path of same radius.......$MeV$

A electron experiences a force $\left( {4.0\,\hat i + 3.0\,\hat j} \right)\times 10^{-13} N$ in a uniform magnetic field when its velocity is $2.5\,\hat k \times \,{10^7} ms^{-1}$. When the velocity is redirected and becomes $\left( {1.5\,\hat i - 2.0\,\hat j} \right) \times {10^7}$, the magnetic force of the electron is zero. The magnetic field $\vec B$ is :

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 (charge $q$ $coulomb$) enters a magnetic field of $H$ $weber/{m^2}$ with a velocity of $v\,m/s$ in the same direction as that of the field the force on the electron is

If two streams of protons move parallel to each other in the same direction, then they

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is $6: 5$ and their respective masses ratio is $9: 4$. Then, the ratio of their charges will be.