Question

Consider the motion of a particle with mass m and electric charge ${\mathbf{q}}_{\mathbf{e}}$in the field of a (hypothetical) stationary magnetic monopole ${\mathbf{q}}_{\mathbf{m}}$at the origin:

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}}\frac{{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}$

(a) Find the acceleration of ${\mathbf{q}}_{\mathbf{e}}$, expressing your answer in terms of localid="1657533955352" $\mathbf{q}$, ${\mathbf{q}}_{\mathbf{m}}$, $\mathbf{m}$, r (the position of the particle), and $\mathbf{v}$(its velocity).

(b) Show that the speed $\mathbf{v}\mathbf{=}\left|v\right|$is a constant of the motion.

(c) Show that the vector quantity

$\mathbf{Q}\mathbf{=}\mathbf{m}\mathbf{(}\mathbf{r}\mathbf{\times }\mathbf{v}\mathbf{)}\mathbf{-}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{4}}^{\mathbf{\pi }}}\hat{\mathbf{r}}$

is a constant of the motion. [Hint: differentiate it with respect to time, and prove-using the equation of motion from (a)-that the derivative is zero.]

(d) Choosing spherical coordinates localid="1657534066650" $\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{,}\varphi \mathbf{)}$, with the polar (z) axis along $\mathbf{Q}$,

(i) calculate , localid="1657533121591" $\mathbf{Q}‐\hat{\varphi }$and show that $\mathbf{\theta }$is a constant of the motion (so ${\mathbf{q}}_{\mathbf{e}}$moves on the surface of a cone-something Poincare first discovered in 1896${\mathbf{)}}^{\mathbf{24}}$;

(ii) calculate $\mathbf{Q}‐\hat{\mathrm{r}}$, and show that the magnitude of $\mathbf{Q}$is

$\mathbf{Q}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\left|\frac{q_e{q}_m}{\mathrm{cos\theta }}\right|$;

(iii) calculate $\mathbf{Q}‐\hat{\mathrm{\theta }}$, show that

$\frac{\mathbf{d\varphi }}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}$,

and determine the constant k .

(e) By expressing ${\mathbf{v}}^{\mathbf{2}}$in spherical coordinates, obtain the equation for the trajectory, in the form

$\frac{\mathbf{dr}}{\mathbf{d}\varphi }=\mathbf{f}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$

(that is: determine the function )$\mathbf{f}\left(\mathbf{r}\right))$.

(t) Solve this equation for .$\mathbf{r}\left(\varphi \right)$

Short answer

(a) The acceleration of $q_e$is $\frac{{\mu }_0{q}_m}{4m\pi r^2}qv\times \hat{r}$.

(b) The speed $v=\left|v\right|$is a constant of the motion.

(c) The vector quantity is a constant of the motion.

(d)

(i) $Q‐\hat{\varphi }$is and $\theta$is a constant of the motion.

(ii) The value of $Q‐\hat{r}$is $Q\cos \theta$and the magnitude of $\mathrm{Q}$is proved.

(iii) The value of $Q‐\hat{\theta }$is $-Q\sin \theta$and $\frac{d\varphi }{dt}=\frac{k}{r^2}$is proved.

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

(e) The function f (r) is $r\sqrt{{\left(\frac{rv}{k}\right)}^2}-{\sin }^2\theta$.

(f) $r\left(\varphi \right)$is $\frac{A}{\cos \left(ar\cos {\displaystyle \frac{A}{r_0}}+\sin \theta (\varphi -{\varphi }_0\right)}$.

Step-by-step solution

Identification of the given data

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 $q_m$,

Significance of the magnetostatics

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’ magnetic analogue where there are stationary charges.

(a) Determination of the acceleration

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

F= ma

…(i)

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:

$F=\frac{{\mu }_0{q}_m}{4\pi {\mathrm{r}}^2}qv\times \hat{r}$

…(ii)

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’s charge, role="math" localid="1657535676962" ${\mu }_0$is the permeability constant $\hat{r}$and is the position vector.

Equalling the equation (i) and (ii).

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

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

(b) Determination of the speed

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

$a‐v=0$

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.

(c) Determination of the prove that the vector quantity is a constant of the motion

The equation of the vector quantity is expressed as:

$Q=m(r\times v)-\frac{{\mu }_0{q}_e{q}_m}{4\pi }\hat{r}$

…(iii)

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’s electric charge, ${\mu }_0$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.

$$\begin{gathered} \frac{dQ}{dt}=m(v\times v+r\times a)-\frac{{\mu }_0{q}_e{q}_m}{4\mathrm{\pi }}\frac{vr-rr}{r^2} \\ =mr\times \left(\frac{{\mu }_0{q}_e{q}_m}{4m\pi {\mathrm{r}}^2}v\times \hat{r}\right)-\frac{{\mu }_0{q}_e{q}_m}{4\pi {\mathrm{r}}^2}(vr-rr) \\ =\frac{{\mu }_0{q}_e{q}_m}{4\pi r^{\mathit{2}}}(r\times (v\times \hat{r}\left)\right)-\frac{{\mu }_0{q}_e{q}_m}{4\pi r^2}(vr-rr) \end{gathered}$$ …(iv)

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

$$\begin{gathered} r\times (v\times r)=v(r‐\hat{r})-\hat{r}(r‐v) \\ =vr-\hat{r}r\dot{r} \\ =vr-r\hat{r} \end{gathered}$$

Substitute the above value in the equation (iv).

$\frac{dQ}{dt}=\frac{{\mu }_0{q}_e{q}_m}{4\pi r^2}(vr-\dot{r}r)-\frac{{\mu }_0{q}_e{q}_m}{4\pi r^{\mathit{2}}}(vr-\dot{r}r)$

= 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.

(d) (i) Determination of Q.ϕ¯

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

$$\begin{gathered} r\times v=r\times (\dot{r}\hat{r}+r\dot{\theta }\hat{\theta }+r\sin \theta \dot{\varphi }\hat{\varphi }) \\ =r^2\dot{\theta }\hat{\varphi }-r^2\sin \theta \hat{\theta }\dot{\varphi } \end{gathered}$$

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

$Q=m{r}^2\dot{\theta }\hat{\varphi }-m{r}^2\sin \theta \hat{\theta }\dot{\varphi }-\frac{{\mu }_0{q}_e{q}_m}{4\mathrm{\pi }}\hat{r}$

…(v)

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

$$\begin{gathered} Q‐\hat{\varphi }=m{r}^2\dot{\theta }\dot{\varphi }\hat{\varphi }-\frac{{\mu }_0{q}_e{q}_m}{4\mathrm{\pi }}\hat{r}\hat{\varphi } \\ =m{r}^2\dot{\theta } \\ =0 \end{gathered}$$

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

Thus, localid="1657620427016" $Q.\hat{\varphi }$is 0 and $\theta$is a constant of the motion.

(d) (ii) Determination of Q.r^

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

$$\begin{gathered} Q‐\hat{r}=m{r}^2\dot{\theta }\hat{\varphi }\hat{r}-m{r}^2\sin \theta \hat{\theta }\dot{\varphi }\hat{r}-\frac{{\mu }_0{q}_e{q}_m}{4\mathrm{\pi }}\hat{r}\hat{r} \\ =Q\cos \theta \end{gathered}$$

The above equation can also be expressed as:

$$\begin{gathered} Q\cos \theta =\frac{{\mu }_0{q}_e{q}_m}{4\pi } \\ Q=\frac{{\mu }_0{q}_e{q}_m}{4\pi \cos \theta } \end{gathered}$$

Thus, the value of $Q‐\hat{r}$is $Q\cos \theta$and the magnitude of $Q$is proved.

(d) (iii) Determination of Q.θ^

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

$$\begin{gathered} Q.\hat{\theta }=m{r}^2\dot{\theta }\hat{\varphi }\hat{\theta }-m{r}^2\sin \theta \hat{\theta }-\frac{{\mu }_0{q}_e{q}_m}{4\mathrm{\pi }}\hat{r}\hat{\theta } \\ =-Q\sin \theta \end{gathered}$$

The above equation can also be expressed as:

$$\begin{gathered} -Q\sin \theta =-m{r}^2\sin \theta \dot{\varphi } \\ \dot{\phi }=\frac{Q}{m{r}^2} \end{gathered}$$

…(vi)

As the dot product $Q‐\hat{\theta }$is , $-Q\sin \theta$then the constant is $k$expressed as:

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

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

$$\begin{gathered} \dot{\phi }\frac{k}{r^2} \\ \frac{d\varphi }{dt}=\frac{k}{r^2} \end{gathered}$$

Thus, $Q‐\hat{\theta }$the value of$-Q\sin \theta$ is and $\frac{d\varphi }{dt}=\frac{k}{r^2}$is proved.

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

(e) Determination of the function f (r)

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

$v^2=\dot{r}+r^2{\sin }^2\theta {\dot{\varphi }}^2+r^2{\dot{\theta }}^2$

Substitute 0 for $\dot{\theta }$and $\frac{Q}{m{r}^2}$for in the above equation.

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

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

$$\begin{gathered} \dot{r}=\frac{dr}{dt} \\ =\frac{dr}{d\varphi }\frac{d\varphi }{dt} \\ =\frac{dr}{d\varphi }\dot{\varphi } \\ \frac{\dot{r}}{\dot{\varphi }}=\frac{dr}{d\varphi } \end{gathered}$$

…(vii)

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

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

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

$$\begin{gathered} \frac{dr}{d\varphi }=\frac{rm}{Q}\sqrt{r^2{v}^2-C^2} \\ =r\sqrt{{\left(\frac{rv}{k}\right)}^2-{\sin }^2\theta } \\ =f\left(r\right) \end{gathered}$$ …(viii)

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

(f) Solving the equation

The equation (viii) can be recalled as:

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

Integrating the above equation with the limits.

${\int }_{{\varphi }_o}^{\varphi }d\varphi ={\int }_{r_o}^r\frac{dr}{r\sqrt{{\left({\displaystyle \frac{rv}{k}}\right)}^2-{\sin }^2\theta }}$

$\varphi -{\varphi }_0\frac{k}{v}{\int }_{r_o}^r\frac{dr}{r\sqrt{r^2-A^2}}$

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

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

Hence, further as:

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

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

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