Question

Suppose an electron of mass m and charge $-e$ revolves round a nucleus of mass $M$ $(>>m)$ and charge $+Ze$ along a circular orbit of radius $r$ with a speed $u$. A uniform magnetic field $B_o$ is switched on along the normal and out of the plane of the orbit as shown in the Figure. Select TRUE statement/s from the following:

• A. when the magnetic field is turned on, the electron speeds up, keeping the radius of the orbit same
• B. the change in angular velocity, as the magnetic field is turned on, will be $e$$B_o$/$2m$
• C. when the magnetic field is turned on, the electron slows down, keeping the radius of the orbit same
• D. the change in kinetic energy of electron is due to the work done by the magnetic field

Answer 1

Answer: (A), (B)

The correct options are
A when the magnetic field is turned on, the electron speeds up, keeping the radius of the orbit same
B the change in angular velocity, as the magnetic field is turned on, will be $e$$B_o$/$2m$

Let u be the speed of electron when the magnetic field $B_0$ is put off and u be the speed when it is put on. (Keeping radius same)
$\Rightarrow \frac{Z{e}^2}{4\pi {\epsilon }_0{r}^2}=\frac{m{u}^2}{r}$
$\Rightarrow \frac{Z{e}^2}{4\pi {\epsilon }_0{r}^2}+e{u}^{\prime }{B}_0=\frac{m{u^{\prime }}^2}{r}$
From the above two equations we can see that
$\Rightarrow \frac{m{u}^2}{r}+e{u}^{\prime }{B}_0=\frac{m{u^{\prime }}^2}{r}$
Therefore the electron speeds up.
$\Rightarrow \frac{m}{r}(u^{\prime 2}-u^2)=e{u}^{\prime }{B}_0$
$\Rightarrow (u^{\prime }-u)(u^{\prime }+u)=\frac{e{B}_{0}r}{m}{u}^{\prime }$
For $u^{\prime }>>u,\Delta u=\frac{e{B}_{0}r}{2m}$
$\Rightarrow \Delta \omega =\frac{e{B}_0}{2m}$