91影视

The given data is listed below as:

• The mass of the particle is, m
• The electric charge of the particle is, $q_e$
• The stationary magnetic monopole is $q_m$,

Magnetostatics is described as the study of the magnetic fields in the systems where the currents remain steady. However, magnetostatics is described as the electrostatics鈥� magnetic analogue where there are stationary charges.

The equation of the force of the particle is expressed as:

F= ma

鈥�(颈)

Here, m is the mass of the particle and a is the acceleration of the particle.

The equation of the force of the particle can also be written as:

$F=\frac{{渭}_0{q}_m}{4蟺{\mathrm{r}}^2}qv脳\hat{r}$

鈥�(颈颈)

Here, $q_m$is the stationary magnetic monopole, v is the velocity of the particle, r is the position of the particle, q is the particle鈥檚 charge, role="math" localid="1657535676962" ${渭}_0$is the permeability constant $\hat{r}$and is the position vector.

Equalling the equation (i) and (ii).

role="math" localid="1657536121138" $$\begin{gathered} ma=\frac{{渭}_0{q}_m}{4蟺r^{\mathit{2}}}qv脳\hat{r} \\ a=\frac{{渭}_0{q}_m}{4m蟺r^{\mathit{2}}}qv脳\hat{r} \end{gathered}$$

Thus, the acceleration of $q_e$is $\frac{{渭}_0{q}_m}{4m蟺r^{\mathit{2}}}qv脳\hat{r}$.

The magnitude of the velocity of the particle is conserved. Hence, the equation of the speed of the particle is expressed as:

$a鈥�v=0$

Here, is the acceleration of the particle andv is the velocity of the particle.

The above equation can also be written as:

$$\begin{gathered} a.v=\frac{dv}{dt}v \\ =\frac{1}{2}\frac{d}{dt}\left({\left|v\right|}^2\right) \\ =0 \end{gathered}$$

Hence, the equation is proved.

Thus, the speed $v=\left|v\right|$is a constant of the motion.

The equation of the vector quantity is expressed as:

$Q=m(r脳v)-\frac{{渭}_0{q}_e{q}_m}{4蟺}\hat{r}$

鈥�(颈颈颈)

Here, $q_m$is the stationary magnetic monopole, v is the velocity of the particle, r is the position of the particle, $q_e$is the particle鈥檚 electric charge, ${渭}_0$is the permeability constant, m is the mass of the particle and is $\hat{r}$the position vector.

Differentiating the above equation with respect to time.

$$\begin{gathered} \frac{dQ}{dt}=m(v脳v+r脳a)-\frac{{渭}_0{q}_e{q}_m}{4\mathrm{蟺}}\frac{vr-rr}{r^2} \\ =mr脳\left(\frac{{渭}_0{q}_e{q}_m}{4m蟺{\mathrm{r}}^2}v脳\hat{r}\right)-\frac{{渭}_0{q}_e{q}_m}{4蟺{\mathrm{r}}^2}(vr-rr) \\ =\frac{{渭}_0{q}_e{q}_m}{4蟺r^{\mathit{2}}}(r脳(v脳\hat{r}\left)\right)-\frac{{渭}_0{q}_e{q}_m}{4蟺r^2}(vr-rr) \end{gathered}$$ 鈥�(颈惫)

For the triple product, the BAC-CAB rule is used.

$$\begin{gathered} r脳(v脳r)=v(r鈥�\hat{r})-\hat{r}(r鈥�v) \\ =vr-\hat{r}r\stackrel{藱}{r} \\ =vr-r\hat{r} \end{gathered}$$

Substitute the above value in the equation (iv).

$\frac{dQ}{dt}=\frac{{渭}_0{q}_e{q}_m}{4蟺r^2}(vr-\stackrel{藱}{r}r)-\frac{{渭}_0{q}_e{q}_m}{4蟺r^{\mathit{2}}}(vr-\stackrel{藱}{r}r)$

= 0

As the derivate is zero, hence it can be stated that the vector quantity is a constant of the motion.

Thus, the vector quantity is a constant of the motion.

The equation of the cross product of the particle鈥檚 position and the velocity is expressed as:

$$\begin{gathered} r脳v=r脳(\stackrel{藱}{r}\hat{r}+r\stackrel{藱}{胃}\widehat{胃}+r\sin 胃\stackrel{藱}{蠒}\widehat{蠒}) \\ =r^2\stackrel{藱}{胃}\widehat{蠒}-r^2\sin 胃\widehat{胃}\stackrel{藱}{蠒} \end{gathered}$$

Substitute the values from the above equation in the equation (iii).

$Q=m{r}^2\stackrel{藱}{胃}\widehat{蠒}-m{r}^2\sin 胃\widehat{胃}\stackrel{藱}{蠒}-\frac{{渭}_0{q}_e{q}_m}{4\mathrm{蟺}}\hat{r}$

鈥�(惫)

The dot product of the above equation with is expressed as:

$$\begin{gathered} Q鈥�\widehat{蠒}=m{r}^2\stackrel{藱}{胃}\stackrel{藱}{蠒}\widehat{蠒}-\frac{{渭}_0{q}_e{q}_m}{4\mathrm{蟺}}\hat{r}\widehat{蠒} \\ =m{r}^2\stackrel{藱}{胃} \\ =0 \end{gathered}$$

As both the variables are mutually orthogonal. Hence, localid="1657620419227" $\stackrel{\mathit{藱}}{胃}=0$and it is constant of motion.

Thus, localid="1657620427016" $Q.\widehat{蠒}$is 0 and $胃$is a constant of the motion.

The dot product of the equation (v) with $\hat{r}$is expressed as:

$$\begin{gathered} Q鈥�\hat{r}=m{r}^2\stackrel{藱}{胃}\widehat{蠒}\hat{r}-m{r}^2\sin 胃\widehat{胃}\stackrel{藱}{蠒}\hat{r}-\frac{{渭}_0{q}_e{q}_m}{4\mathrm{蟺}}\hat{r}\hat{r} \\ =Q\cos 胃 \end{gathered}$$

The above equation can also be expressed as:

$$\begin{gathered} Q\cos 胃=\frac{{渭}_0{q}_e{q}_m}{4蟺} \\ Q=\frac{{渭}_0{q}_e{q}_m}{4蟺\cos 胃} \end{gathered}$$

Thus, the value of $Q鈥�\hat{r}$is $Q\cos 胃$and the magnitude of $Q$is proved.

The dot product of the equation (v) with is expressed as:

$$\begin{gathered} Q.\widehat{胃}=m{r}^2\stackrel{藱}{胃}\widehat{蠒}\widehat{胃}-m{r}^2\sin 胃\widehat{胃}-\frac{{渭}_0{q}_e{q}_m}{4\mathrm{蟺}}\hat{r}\widehat{胃} \\ =-Q\sin 胃 \end{gathered}$$

The above equation can also be expressed as:

$$\begin{gathered} -Q\sin 胃=-m{r}^2\sin 胃\stackrel{藱}{蠒} \\ \stackrel{藱}{蠁}=\frac{Q}{m{r}^2} \end{gathered}$$

鈥�(惫颈)

As the dot product $Q鈥�\widehat{胃}$is , $-Q\sin 胃$then the constant is $k$expressed as:

$k=\frac{Q}{m}$

Substitute the value of the above equation in equation (vi).

$$\begin{gathered} \stackrel{藱}{蠁}\frac{k}{r^2} \\ \frac{d蠒}{dt}=\frac{k}{r^2} \end{gathered}$$

Thus, $Q鈥�\widehat{胃}$the value of$-Q\sin 胃$ is and $\frac{d蠒}{dt}=\frac{k}{r^2}$is proved.

The constant $k$is $\frac{Q}{m}$.

The equation of expressing the square of the velocity in the spherical coordinates is expressed as:

$v^2=\stackrel{藱}{r}+r^2{\sin }^2胃{\stackrel{藱}{蠒}}^2+r^2{\stackrel{藱}{胃}}^2$

Substitute 0 for $\stackrel{藱}{胃}$and $\frac{Q}{m{r}^2}$for in the above equation.

role="math" localid="1657618825571" $$\begin{gathered} v^2={\stackrel{藱}{r}}^2+r^2{\sin }^2胃\frac{Q_2}{m^2{r}^4} \\ ={\stackrel{藱}{r}}^2+\frac{C^2}{r^2} \\ {C}^2={\sin }^2胃\frac{Q^2}{m^2} \end{gathered}$$

The chain rule has been applied in the differential equation of the curve.

$$\begin{gathered} \stackrel{藱}{r}=\frac{dr}{dt} \\ =\frac{dr}{d蠒}\frac{d蠒}{dt} \\ =\frac{dr}{d蠒}\stackrel{藱}{蠒} \\ \frac{\stackrel{藱}{r}}{\stackrel{藱}{蠒}}=\frac{dr}{d蠒} \end{gathered}$$

鈥�(惫颈颈)

The equation of the position of the particle is expressed as:

$\stackrel{藱}{r}=\sqrt{v^2-\frac{C^2}{r^2}}$

Substitute $\sqrt{v^2-\frac{C^2}{r^2}}$for $\stackrel{藱}{r}$and$\frac{Q}{m{r}^2}$for $\stackrel{藱}{蠒}$in the equation (vii).

$$\begin{gathered} \frac{dr}{d蠒}=\frac{rm}{Q}\sqrt{r^2{v}^2-C^2} \\ =r\sqrt{{\left(\frac{rv}{k}\right)}^2-{\sin }^2胃} \\ =f\left(r\right) \end{gathered}$$ 鈥�(惫颈颈颈)

Thus, $f\left(r\right)$the function is $r\sqrt{{\left(\frac{rv}{k}\right)}^2-{\sin }^2胃}$.

The equation (viii) can be recalled as:

$$\begin{gathered} \frac{dr}{d蠒}=r\sqrt{{\left(\frac{rv}{k}\right)}^2-{\sin }^2胃} \\ d蠒=\frac{dr}{r\sqrt{{\left({\displaystyle \frac{rv}{k}}\right)}^2-{\sin }^2胃}} \end{gathered}$$

Integrating the above equation with the limits.

${鈭�}_{{蠒}_o}^{蠒}d蠒={鈭�}_{r_o}^r\frac{dr}{r\sqrt{{\left({\displaystyle \frac{rv}{k}}\right)}^2-{\sin }^2胃}}$

$蠒-{蠒}_0\frac{k}{v}{鈭�}_{r_o}^r\frac{dr}{r\sqrt{r^2-A^2}}$

Substitute $\frac{{\sin }^2胃k^2}{v^2}$for in the above equation.

$$\begin{gathered} 蠒-{蠒}_o={{\left[\frac{k}{vA}arcc\mathrm{os}\frac{A}{r}\right]}^r}_{r_o} \\ =\frac{1}{\sin 胃}\left(arc\cos \frac{A}{r}-arc\cos \frac{A}{r_0}\right) \\ arc\cos \frac{A}{r}=arc\cos \frac{A}{r_o}+\sin 胃(蠒-{蠒}_o) \\ \cos \left(arc\cos \frac{A}{r}\right)=\cos \left(arc\cos \frac{A}{r_o}+\sin 胃(蠒-{蠒}_o\right) \end{gathered}$$

Hence, further as:

$$\begin{gathered} \frac{A}{r}=\cos \left(arc\cos \frac{A}{r_0}+\sin 胃(蠒-{蠒}_0\right) \\ r=\frac{A}{\cos \left(arc\cos \frac{A}{r_0}+\sin 胃(蠒-{蠒}_0)\right)} \end{gathered}$$

The term $\frac{A}{r_0}$is the integration constant and it rotates around the orbit.

Thus, $r\left(蠒\right)$is $\frac{A}{\cos \left(arc\cos {\displaystyle \frac{A}{r_o}}+\sin 胃(蠒-{蠒}_o)\right)}$.