Question; An atomic electron (charge q ) circles about the nucleus (charge Q) in an orbit of radius r ; the centripetal acceleration is provided, of course, by the Coulomb attraction of opposite charges. Now a small magnetic field dB is slowly turned on, perpendicular to the plane of the orbit. Show that the increase in kinetic energy, dT , imparted by the induced electric field, is just right to sustain circular motion at the same radius r. (That's why, in my discussion of diamagnetism, I assumed the radius is fixed. See Sect. 6.1.3 and the references cited there.)

The charge of the electron is q .

The charge of the nucleus is Q.

The radius of the orbit is r.

The small magnetic field is db.

The expression to calculate the centripetal force is given as follows.

$F_{cp}=\frac{m{v}^2}{r}$

Here, m is the mass of object, v is the velocity and r is the radius of the circular path.

The expression to calculate the electrostatic force is given as follows.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^2}$

Here,q₁ ,q₂ are the two charges and is the distance between the charges.

$dT=T_1-T$The expression of centripetal force in the absence of the magnetic field is given by,

$F_{cp}=\frac{m{v}^2}{r}$

The coulombs force between the electron and nucleus is given by,

$F={\frac{1}{4\pi \epsilon }}_0\frac{Qq}{r^2}$

The centripetal force is produced due to coulombs attraction force.

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r^2} \\ m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r} \end{gathered}$$

The kinetic energy of the electron is given by,

$T=\frac{1}{2}m{v}^2$

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r}$ for mv² into above equation.

$$\begin{gathered} T=\frac{1}{2}\times \frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r} \\ T=\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r} \end{gathered}$$

The magnetic force of electron when the magnetic force is applied perpendicular to the plane is given by,

$$\begin{gathered} F_B=q\left(\overrightarrow{v_1}\times \overline{dB}\right) \\ {F}_B=q{v}_{1}Bson90° \\ {F}_B=q{v}_{1}B \end{gathered}$$

Here v₁ is the velocity of the electron.

Therefore, centripetal force is provided by the coulomb’s attraction and magnetic force.

$$\begin{gathered} \frac{m{v}_1^2}{r_1}=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r_1^2}+q{v}_{1}B \\ m{v}_1^2=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r_1}+q{v}_1{r}_{1}B \end{gathered}$$

The new kinetic energy of the electron is given by,

$T_1=\frac{1}{2}m{v}_1^2$

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r_1}+q{v}_1{r}_{1}B$for into above equation.

localid="1658312628596" $T_1=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r_1}+q{v}_1{r}_{1}B\right]$ ……. (1)

Let the radius is increased by when the magnetic field is applied.

$$\begin{gathered} r_1=r+dr \\ {r}_1^{-1}={\left(r+dr\right)}^{-1} \\ {r}_1^{-1}=r^{-1}{\left(1+\frac{dr}{r}\right)}^{-1} \end{gathered}$$

Expand the above term ${\left(1+\frac{dr}{r}\right)}^{-1}$by using the binomial expansion.

$r_1^{-1}=r^{-1}\left(1-\frac{dr}{r}\right)$

Let the velocity is increased by when the magnetic field is applied.

$v_1=v+dv$

Substitute $r^{-1}\left(1-\frac{dr}{r}\right)$for $\frac{1}{r}$and r+ d r and for d into equation (1).

$$\begin{gathered} T_1=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}qQ{r}^{-1}\left(1-\frac{dr}{r}\right)+q\left(v+dv\right)\left(r+dr\right)dB\right] \\ {T}_1=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}qQ{r}^{-1}\left(1-\frac{dr}{r}\right)+qvrdB\right] \end{gathered}$$

The increase in the kinetic energy is given by,

Substitute $\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}qQ{r}^{-1}\left(1-\frac{dr}{r}\right)+qvrdB\right]$for T₁ and $\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r}$ for Tinto above equation.

$$\begin{gathered} dT=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}qQ{r}^{-1}\left(1-\frac{dr}{r}\right)+qvrdB\right]-\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r} \\ dT=\frac{1}{2}\left[\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r}\left(1-\frac{dr}{r}\right)+qvrdB\right]-\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r} \\ dT=\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r}\left(1-\frac{dr}{r}-1\right)+\frac{qvr}{2}dB \\ dT=-\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r}\frac{dr}{r}+\frac{qvr}{2}dB \end{gathered}$$

The induced electric field is given by,

$E=\frac{r}{2}\frac{dB}{dt}$

The induced electrical force is given by,

$m\frac{dv}{dt}=qE$

Substitute$\frac{r}{2}\frac{dB}{dt}$for E into above equation.

$$\begin{gathered} m\frac{dv}{dt}=q\left(\frac{r}{2}\frac{dB}{dt}\right) \\ mdv=\frac{qr}{2}dB \end{gathered}$$

Multiply the above equation by V both the side.

$mvdv=\frac{vqr}{2}dB$

Therefore, increase in kinetic energy is given by,

$$\begin{gathered} dT=d\left(\frac{1}{2}m{v}^2\right) \\ =\frac{1}{2}md\left(v^2\right) \\ =\frac{1}{2}m\left(2vdv\right) \\ =mvdv \end{gathered}$$

Substitute $\frac{vqr}{2}dB$ for $mvdv$into above equation.

$dT=\frac{vqr}{2}dB$

Substitute$-\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r}\frac{dr}{r}+\frac{qvr}{2}dB$for dT into above equation.

$$\begin{gathered} -\frac{1}{8\pi {\epsilon }_0}\frac{qQ}{r}\frac{dr}{r}+\frac{qvr}{2}dB=\frac{vqr}{2}dB \\ \frac{1}{2}\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{r^2}dr=0 \\ dr=0 \end{gathered}$$

This implies there is no charge in radius, that is $r_1=r$.

Hence it is showed that the increase in kinetic energy, imparted by the induced electric field, is just right to sustain circular motion at the same radius .