$A$ proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius

An electron is moving in the north direction. It experiences a force in the vertically upward direction. The magnetic field at the position of the electron is in the direction of

An electron moving with a velocity $\vec{V_1} = 2\,\hat{i}\,\text{m/s}$ at a point in a magnetic field experiences a force $\vec{F_1} = -2\hat{j}\,\text{N}$. If the electron is moving with a velocity $\vec{V_2} = 2\,\hat{j}\,\text{m/s}$ at the same point,it experiences a force $\vec{F_2} = +2\,\hat{i}\,\text{N}$. The force the electron would experience if it were moving with a velocity $\vec{V_3} = 2\hat{k}\,\text{m/s}$ at the same point is

$A$ neutron,a proton,an electron,and an $\alpha$-particle enter a region of uniform magnetic field with the same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labeled in the figure. Which track does the electron follow?

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 magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of:

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?