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 electron remains undeviated. The speed of electron if the electric field is removed, and the radius of electron path will be respectively

An electron is moving with a speed of ${10^8}\,m/\sec $ perpendicular to a uniform magnetic field of intensity $B$. Suddenly intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that

An electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic field perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the $x-$ axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$ . What should be magnetic field such that the charge particle just escapes the wall

When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with  a speed $v_0$ it moves with an initial acceleration $3a_0$ toward west. The electric and magnetic fields in the room are