An ion beam of specific charge $5 \times 10^7$ $coulomb/kg$ enter a uniform magnetic field of $4 \times 10^{-2}\, tesla$ with a velocity $2 \times 10^5\, m/s$ perpendicularly. The radius of the circular path of ions in meter will be

An electron (mass = $9.0 × $${10^{ - 31}}$ $kg$ and charge =$1.6 \times {10^{ - 19}}$ $coulomb$) is moving in a circular orbit in a magnetic field of $1.0 \times {10^{ - 4}}\,weber/{m^2}.$ Its period of revolution is

A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled, then the radius of the circular path will be.....$cm$

The magnetic field is uniform for $y>0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y<0$. A proton denoted by filled circle leaves $y=0$ in the $-y$-direction with some speed as shown below.Which of the following best denotes the trajectory of the proton?

The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to

Show that a force that does no work must be a velocity dependent force.