The figure shows three situations when an electron with velocity $\vec v$ travels through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

An electron is moving in a circle of radius $r$ in a magnetic field $B$. Suddenly,the field is reduced to $B/2$. The radius of the circular path is

In a chamber,a uniform magnetic field of $6.5 \;G \left(1 \;G = 10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $\left(e = 1.6 \times 10^{-19} \;C, m_{e} = 9.1 \times 10^{-31} \;kg \right)$

An electron of mass $m$ and charge $e$ is projected normally from the surface of a sphere of radius $a$ with speed $v_0$ into a uniform magnetic field $B$ perpendicular to the plane of the paper. The centre of the sphere is at a distance $b$ from a wall. If the electron strikes the wall symmetrically with respect to the $x$-axis,what should be the magnetic field $B$ such that the charged particle just touches the wall?

An electron is moving along the $+x$ direction. To make it move along an anticlockwise circular path in the $x-y$ plane,the magnetic field must be applied along:

An electron projected perpendicular to a uniform magnetic field $B$ moves in a circle. If Bohr's quantization is applicable,then the radius of the electronic orbit in the first excited state is $:$