13.1: Chapter 1

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Contributed by John F. Cochran and Bretislav Heinrich
Professors (Physics) at Simon Fraser University

Problem (1.1).

Two charges, each q=+1.6x10^-19 Coulombs, are located at (0,0,a) and at (0,0,-a) where a=1.0x10^-9 meters.

(a) Calculate the electric field at the origin (0,0,0).

(Answ: the field is zero.)

(b) Calculate the electric field at (a,a,a).

(Answ: E= (6.07,6.07,1.96)x10⁸ Volts/m.)

(c) An electron, q=-1.6x10^-19 Coulombs, flies through the point (a,a,a) with the velocity v= v₀(1,2,3) where v₀= 10⁵ m/sec. What forces are exerted on the electron due to the two stationary charges?

(Answ: F= qE= (-9.71,-9.71,-3.14)x10^-11 Newtons. There is no magnetic force.)

Problem (1.2).

At a certain moment a moving proton, q=+1.6x10^-19 Coulombs, is located at (0,0,a) with velocity components v₀(1,1,0) where a=10^-9 m. and v₀=10⁵ m/sec. At the same moment a moving electron, q=-1.6x10^-19 Coulombs, is located at (a,a,a) with velocity components (0,10⁶,0) m/sec.

(a) Calculate the electric and magnetic fields at the position of the electron due to the proton.

(Answ: E= (E₀,E₀,0) where E₀= 5.09x10⁸ V/m. and B= (0,0,0) because v_pxE =0.)

(b) Calculate the force on the electron due to the electric field of the proton.

(Answ: F=(-F₀,-F₀,0) where F₀= |q|E₀ =8.14x10^-11 N.)

(Answ: F= v_electronxB = 0 N.)

(d) Calculate the electric and magnetic forces on the proton due to the fields generated by the electron.

Answ: The electric field at the position of the proton, R=(0,0,a), due to the electron at r=(a,a,a) is given by

\[\mathbf{E}=\frac{1}{4 \pi \varepsilon_{0}} \quad\left(-1.6 \times 10^{-19}\right) \frac{\rho}{\rho^{3}}, \nonumber\]

where ρ= R-r = (-a,-a,0)= - a(1,1,0), where a= 10^-9 m.

Therefore

\[\mathbf{E}=\left(5 \cdot 09 \times 10^{8}\right)(1,1,0). \nonumber\]

The magnetic field at the position of the proton due to the motion of the electron is given by c²B= vxE, where the velocity of the electron is v= 10⁶(0,1,0) m/sec. c²B= (5.09x10¹⁴)(0,0,-1) so B= (0.566x10^-2)(0,0,-1) Teslas.

The force on the proton due to the electric field is F_E= 8.15x10^-11(1,1,0) N. The force on the proton due to the magnetic field is F_M= q(v_pxB) = 0.906x10^-16(-1,1,0) N.

Problem (1.3).

A particle having a velocity V=v₁u_x carries a charge q₁ C and is located at the origin. A second particle, charge q₂, is located at r= au_x + bu_y + cu_z, and it has a velocity V₂=v_yu_y.

(a) Show that the force on charge #2 due to the magnetic field generated by charge #1 is \(\mathbf{F}_{21}=\frac{\mu_{0}}{4 \pi} \frac{\text{q}_{1} \text{q}_{2}}{\text{r}^{3}} \text{bv}_{1} \text{v}_{\text{y}} \mathbf{u}_{\text{x}}\).

(b) Show that the force on charge #1 due to the magnetic field generated by charge #2 is \(\mathbf{F}_{12}=-\frac{\mu_{0}}{4 \pi} \frac{\text{q}_{1} \text{q}_{2}}{\text{r}^{3}} \text{av}_{1 \text{V} \text{y}} \mathbf{u}_{\text{y}}\). Notice that F₂₁ does not equal -F₁₂ so that Newton's law of the equality of forces of action and reaction is not obeyed in this case.

Answer (1.3).

(a) The electric field at the position of particle #2 due to particle #1 is

\[\mathbf{E}_\mathbf{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text{q}_{1}}{\text{r}^{3}}(\text{a}, \text{b}, \text{c}). \nonumber\]

The magnetic field at the position of particle #2 due to the motion of particle #1 is given by

\[c^{2} \mathbf{B}_\mathbf{21}=\mathbf{v}_{1} \times \mathbf{E}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\text{q}_{1} \text{v}_{1}}{\text{r}^{3}}(0,-\text{c}, \text{b}), \nonumber \]

\[\mathbf{B}_\mathbf{21}=\frac{\mu_{0}}{4 \pi} \frac{\text{q}_{1} \text{v}_{1}}{\text{r}^{3}}(0,-\text{c}, \text{b}). \nonumber\]

The magnetic force on particle #2 due to its motion is

\[\mathbf{F}_\mathbf{2 M}=q_{2}\left(\mathbf{v}_\mathbf{2} \times \mathbf{B}_\mathbf{21}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{1} q_{2} v_{1} v_{y}}{r^{3}} \quad(b, 0,0).\nonumber\]

(b) The electric field at the position of particle #1 due to particle #2 is

\[\mathbf{E}_\mathbf{12}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{\text{q} 2}{\text{r}^{3}}(\text{a}, \text{b}, \text{c}). \nonumber\]

The magnetic field at the position of particle #1 due to the motion of particle #2 is given by

\[c^{2} \mathbf{B}_\mathbf{12}=\mathbf{v}_{2} \times \mathbf{E}_\mathbf{12}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{\text{q}_{2} \text{v}_{\text{y}}}{\text{r}^{3}}(\text{c}, 0,-\text{a}), \nonumber\]

\[\mathbf{B}_\mathbf{12}=\frac{\mu_{0}}{4 \pi} \frac{\text{q}_{2} \text{v}_{\text{y}}}{\text{r}^{3}}(\text{c}, 0,-\text{a}). \nonumber\]

The magnetic force on particle #1 due to its motion is

\[\mathbf{F}_\mathbf{1M}=\text{q}_{2}\left(\mathbf{v}_\mathbf{1} \times \mathbf{B}_\mathbf{12}\right)=-\frac{\mu_{0}}{4 \pi} \frac{\text{q}_{1} \text{q}_{2} \text{v}_{1} \text{v}_{\text{y}}}{\text{r}^{3}}(0, \text{a}, 0). \nonumber\]

Problem (1.4).

An electron carries a magnetic moment of |m₀|=9.27x10^-24Joules/Tesla= 1 Bohr magneton. Suppose that this magnetic moment is oriented along the z-axis as shown in the figure.

prob 1.4.PNG

(a) At what angle θ is the field measured by an observer at P a maximum?

(Answ: θ= ±\(\pi\)/2.)

(b) If r= 1 micron (10^-6m.) what is the magnitude and direction of this maximum field?

(Answ: |B_max|= 18.54x10^-13 Teslas directed along +z).

(Answ: |B_min|= 9.27x10-13 Teslas directed along -z. The observer is at θ=0 or \(\pi\).)

Answer

\(\mathbf{B}=\frac{\mu_{0}}{4 \pi}\left(3 \frac{m_{0} z \mathbf{r}}{r^{5}}-\frac{m_{0} \mathbf{u}_\mathbf{z}}{r^{3}}\right)\) therefore \(\text{B}_{\text{x}}=\frac{\mu_{0} \text{m}_{0}}{4 \pi} \frac{3 \text{x} \text{z}}{\text{r}^{5}}, \quad \text{B}_{\text{y}}=\frac{\mu_{0} \text{m}_{0}}{4 \pi} \frac{3 \text{yz}}{\text{r}^{5}}, \quad \text{B}_{z}=\frac{\mu_{0} \text{m}_{0}}{4 \pi}\left(\frac{3 \text{z}^{2}}{\text{r}^{5}}-\frac{1}{\text{r}^{3}}\right).\)

\(\text{B}^{2}=\left(\frac{\mu_{0} \text{m}_{0}}{4 \pi \text{r}^{3}} \quad\left(1+\frac{3 \text{z}^{2}}{\text{r}^{2}}\right)\right)\) so B² is a maximum at x=0, y=0, z=r. B² is a minimum at z=0. \(B_{\min }=\frac{\mu_{0} m_{0}}{4 \pi} \frac{1}{r^{3}}. \) \(B_{\max }=2 B_{\min }\).

Problem (1.5).

The energy of interaction between two magnetic dipoles is given by - m₁•B₂ or by - B₁•m₂ where B₁ is the field generated at the position of dipole #2 by dipole #1, and B₂ is the field at dipole #1 generated by dipole #2. Let these two magnetic dipoles be separated by a constant distance R= 10^-6m (1 µm).

(a) Assume that the two dipoles are forced to remain parallel as shown in the figure. At what angle θ is the interaction energy a minimum? What is this minimum energy?

prob 1.5.PNG

(Answ: θ= ± \(\pi\)/2, \(\text{U}_{\min }=-2 \frac{\mu_{0} \text{m}_{1} \text{m}_{2}}{4 \pi \text{R}^{3}}\).

(b) Assume that θ=0 in the figure, but that the two dipoles are free to rotate in the x-y plane. Let m₁|_x = m₁cos\(\alpha_{1}\) and m₁|_y = m₁sin\(\alpha_{1}\). Similarly let m₂|_x = m₂cos\(\alpha_{2}\) and m₂|_y = m₂sin\(\alpha_{2}\). What will be the minimum energy configuration, and what will be the minimum energy?

(Answ: \(\alpha_{1}\) = \(\alpha_{2}\) = 0 or \(\pi\). \(\text{U}_{\text{min}}=-2 \frac{\mu_{0} \text{m}_{1} \text{m}_{2}}{4 \pi \text{R}^{3}}\).)

Answer

(a) \(\mathbf{B}_\mathbf{1}=\frac{\mu_{0}}{4 \pi}\left(\frac{3\left(\mathbf{m}_\mathbf{1} • \mathbf{r}\right) \mathbf{r}}{\text{r}^{5}}-\frac{\mathbf{m}_\mathbf{1}}{\text{r}^{3}}\right), \quad \text{x}=\operatorname{Rcos} \theta, \quad \text{y}=\operatorname{Rsin} \theta, \quad \text{z}=0\)

\(\text{B}_{1 \text{x}}=\frac{\mu_{0}}{4 \pi} \frac{\text{m}_{1}}{\text{R}^{3}} 3 \sin \theta \cos \theta, \quad \text{B}_{1 \text{y}}=\frac{\mu_{0}}{4 \pi} \frac{\text{m}_{1}}{\text{R}^{3}}\left(3 \sin ^{2} \theta-1\right)\),

therefore

\(U=-\mathbf{B}_\mathbf{1} • \mathbf{m}_\mathbf{2}=-\frac{\mu_{0}}{4 \pi} \frac{m_{1} m_{2}}{R^{3}} \quad\left(3 \sin ^{2} \theta-1\right). \nonumber\)

This expression is clearly a minimum when sinθ= ±1.

(b) When θ=0 and r=(R,0,0) one finds

\(\text{B}_{1 \text{x}}=\frac{\mu_{0}}{4 \pi \text{R}^{3}} 2 \text{m}_{1} \cos \alpha_{1}, \quad \text{B}_{1 \text{y}}=-\frac{\mu_{0}}{4 \pi \text{R}^{3}} \text{m}_{1} \sin \alpha_{1}\),

therefore

\(\text{U}=-\mathbf{m}_\mathbf{2} • \mathbf{B}_\mathbf{1}=-\frac{\mu_{0} \text{m}_{1} \text{m}_{2}}{4 \pi \text{R}^{3}} \quad\left(2 \cos \alpha_{1} \cos \alpha_{2}-\sin \alpha_{1} \sin \alpha_{2}\right)\).

This expression clearly has a minimum when cos\(\alpha_{1}\)=cos\(\alpha_{2}\)=1 and sin\(\alpha_{1}\)=sin\(\alpha_{2}\)=0, ie. when \(\alpha_{1}\)=\(\alpha_{2}\)= 0 or \(\pi\).

Problem (1.6)

A proton and an electron are separated by 10^-12 m = d as shown in the sketch.

prob 1.6.PNG

(a) Calculate the strength of the electric field 1 micron (= 10^-6 m) distant from a proton.

(b) Calculate the strength of the electric field a = 1 micron from the above pt dipole at \(\mathbf{r}=\text{a} \hat{\mathbf{u}}_{z}\). What is the direction of this electric field?

(c) Calculate the strength of the electric field a distance a = 1 micron from the dipole at the point \(\mathbf{r}=-\text{a} \hat{\mathbf{u}}_{z}\). What is the direction of this electric field?

(d) Calculate the strength and direction of the electric field at \(\mathbf{r}=\text{a} \hat{\mathbf{u}}_{x}\) where a = 1 micron.

(e) Calculate the strength and direction of the electric field for the above dipole at \(\mathbf{r}=\frac{1}{\sqrt{2}} a\left(\hat{\mathbf{u}}_{x}+\hat{\mathbf{u}}_{z}\right)\) and a= 1 micron.

N.B. \(\hat{\mathbf{u}}_{x}, \hat{\mathbf{u}}_{y}, \hat{\mathbf{u}}_{z}\) are unit vectors along x,y,z.

Answer (1.6)

a) \(|\mathbf{E}|=\frac{1}{4 \pi \varepsilon_{0}} \frac{e}{r^{2}}\)

\(\begin{aligned}
\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \text{e}=1.6 \times 10^{-19} \text{Coulombs} \\
\text{r}=10^{-6} \text{m}
\end{aligned}\)

\(\therefore|\mathbf{E}|=\frac{(9 \times 109)\left(1.6 \times 10^{-19}\right)}{10^{-12}}=\underline{1440} \text{Volts} / \text{m}\).

b) For a point dipole \(\mathbf{E}_{\text{d}}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{\text{r}^{5}}-\frac{\mathbf{p}}{\text{r}^{3}}\right]\)

In this case p and r are both along z and hence parallel

\(\begin{aligned}
\therefore\left|\mathbf{E}_{\text{d}}\right| &=\frac{1}{4 \pi \varepsilon_{0}} \quad\left(\frac{2 \text{p}}{\text{r}^{3}}\right) \\
&=\frac{1}{4 \pi \varepsilon_{0}} \quad\left(\frac{\text{e}}{\text{r}^{2}}\right) \quad\left(\frac{2 \text{d}}{\text{r}}\right) \\
&=\frac{(2)\left(10^{-12}\right)}{10^{-6}}=\left(2 \times 10^{-6}\right) \times \text {part }(\text{a})
\end{aligned}\)

So |E_d| = 2.88 x 10^-3 Volts/m and directed along z.

c) For this part p.r = -e d a

since p = e d \(\hat{\mathbf{u}}_{z}\)

and \(\mathbf{r}=-\text{a} \hat{\mathbf{u}}_{z}\)

therefore \((\mathbf{p} \cdot \mathbf{r}) \mathbf{r}=e a^{2} d \hat{\mathbf{u}}_{z}\)

So \(\frac{(\mathbf{p} \cdot \mathbf{r}) \mathbf{r}}{r^{5}}=\frac{e d}{a^{3}} \hat{\mathbf{u}}_{z}\)

and \(|\mathbf{E} \text{d}|=\frac{1}{4 \pi \varepsilon_{0}}\left|\left(\frac{3 \text{ed}}{\text{a}^{3}} \hat{\mathbf{u}}_{z}-\frac{\text{ed}}{\text{a}^{3}} \hat{\mathbf{u}}_{z}\right)\right|=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \text{ed}}{\text{a}^{3}}\)

or exactly the same as part (b). The electric field is also directed along z, just as in part (b).

(d) Here p.r=0 because p is directed along z whereas r is directed along x.

Therefore

\(\begin{aligned}
\mathbf{E}_{\text{d}}&=\frac{-1}{4 \pi \varepsilon_{0}} \quad \mathbf{p} / \text{r}^{3} \\
&=-\left(\frac{\text{e}}{4 \pi \varepsilon_{0}}\right), \frac{1}{\text{a}^{2}}(\text{d} / \text{a}) \hat{\mathbf{u}}_{\text{z}}
\end{aligned}\)

i.e. directed along –z and half as large as the electric field for a point along the dipole axis and a meters from the dipole.

∴|Ed| = 1.44 x 10^-3 Volts/m

e) \(\mathbf{p}=\text{e} \text{d} \hat{\mathbf{u}}_{z} \quad \therefore \mathbf{p} \cdot \mathbf{r}=\frac{\text{eda}}{\sqrt{2}}\)

\(\mathbf{r}=\frac{a}{\sqrt{2}} \quad\left(\hat{\mathbf{u}}_{x}+\hat{\mathbf{u}}_{z}\right)\)

\(\therefore \quad \mathbf{E} \text{d}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3 \text{ed}}{(\sqrt{2)}(\sqrt{2)}} \frac{\hat{\mathbf{u}}_{\text{x}}+\hat{\mathbf{u}}_{z}}{\text{a}^{3}}-\frac{\text{e} \text{d} \hat{\mathbf{u}}_{z}}{\text{a}^{3}}\right]\)

\(\mathbf{E}_{\text{d}} =\frac{e}{4 \pi \varepsilon_{0}} \frac{\text{d}}{\text{a}^{3}}\left[\frac{3}{2} \quad\left(\hat{\mathbf{u}}_{\text{x}}+\hat{\mathbf{u}}_{z}\right)-\hat{\mathbf{u}}_{z}\right]\)
\(=\frac{\text{e}}{4 \pi \varepsilon_{0}}\left(\frac{1}{\text{a}^{2}}\right)\left(\frac{\text{d}}{2 \text{a}}\right)\left(3 \hat{\mathbf{u}}_{\text{x}}+\hat{\mathbf{u}}_{z}\right)\)

So Ed is directed 18.4° from the xy plane and has the magnitude |Ed| = (1.58 x 10^-6) (1440) = 2.28 x 10^-3 Volts/m.

Problem (1.7).

Show that the magnetic field at the center of a uniformly magnetized sphere containing a small hole at the center is zero. Uniform magnetization means M is constant. Without loss of generality, one can take the magnetization to be directed along the z-axis, ie M= M₀u_z.

(Hint: Add up all the contributions to the field at the center due to volume elements at a distance r from the center. In polar co-ordinates d\(\tau\)= r²dr sinθdθ d\(\phi\), and dm = M₀d\(\tau\)u_z.)

Answer (1.7).

If r= -xu_x - yu_y - zu_z then m•r = - M₀d\(\tau\)z (remember that r is the vector drawn from the magnetic moment to the point of observation).

\(\mathbf{B}=\frac{\mu_{0}}{4 \pi}\left(\frac{3(\mathbf{m} \cdot \mathbf{r}) \mathbf{r}}{r^{5}}-\frac{\mathbf{m}}{r^{3}}\right)\), so that

\(B_{x}=\frac{\mu_{0}}{4 \pi} \frac{M_{0} d \tau}{r^{5}} \quad(3 x z), \quad B_{y}=\frac{\mu_{0}}{4 \pi} \frac{M_{0} d \tau}{r^{5}} \quad(3 y z)\),

\(B_{z}=\frac{\mu_{0}}{4 \pi} \frac{M_{0} d \tau}{r^{5}} \quad\left(2 z^{2}-x^{2}-y^{2}\right)\),

Convert to polar co-ordinates and integrate over θ from 0 to \(\pi\), and over \(\phi\) from 0 to 2\(\pi\). All field components integrate to zero.

Problem (1.8)

The fields generated at the position r from a slowly moving, spinless, point charge are given by \(\mathbf{E}=\frac{q}{4 \pi \varepsilon_{0}} \frac{\mathbf{r}}{r^{3}} \text { and } c \mathbf{B}=\frac{\mathbf{v}}{c} \times \mathbf{E}\). Consider a particle moving in a circular orbit whose position at time t is given by \(\mathbf{a}=a \cos \omega t \hat{\mathbf{u}}_{\mathbf{x}}+\sin \omega t \quad \hat{\mathbf{u}}_{y}\).

(a) Show that the time averaged electric field seen by an observer at \(\mathbf{R}=\text{x} \hat{\mathbf{u}}_{\mathbf{x}}+\text{Y} \hat{\mathbf{u}}_{\mathbf{y}}+\text{z} \hat{\mathbf{u}}_{\mathbf{z}}\) is given by \(<\mathbf{E}_{p}>=\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{\mathbf{R}}{R^{3}}\right)\)to terms of order (a/R)².

(b) Show that to lowest order in (a/R) the magnetic field observed at R is given by

\[<\mathbf{B}_{\text{p}}>=\frac{1}{\text{c}^{2}} \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3z \mathbf{R}}{\text{R}^{5}}-\frac{\hat{\mathbf{u}}_{\text{z}}}{\text{R}^{3}}\right] \quad\left(\frac{\text{qa}^{2} \omega}{2}\right) \nonumber\]

or since \(\mathbf{m}=\left(\frac{\text{qa}^{2} \omega}{2}\right) \hat{\mathbf{u}}_{\text{z}}\) is the magnetic moment (|m|= I\(\pi\)a² where I is the current in Amps)

\[\left\langle\mathbf{B}_{\text{p}}\right\rangle=\frac{\mu_{0}}{4 \pi} \quad\left[\quad 3\left(\frac{\mathbf{m} \cdot \mathbf{R}}{\text{R}^{5}}\right) \mathbf{R}-\frac{\mathbf{m}}{\text{R}^{3}}\right] \nonumber\]

and \(c^{2}=\frac{1}{\varepsilon_{0} \mu_{0}}\).

Answer (1.8)

prob 1.8.PNG

We have r + a = R

∴ r = R - a

where \(\mathbf{R}=X \hat{\mathbf{u}}_{\text{x}}+\text{Y} \hat{\mathbf{u}}_{\text{y}}+\text{z} \hat{\mathbf{u}}_{\text{z}}\)

and \(\mathbf{a}=\text{a} \cos \omega \text{t} \quad \hat{\mathbf{u}}_{\text{x}}+\text{a} \sin \omega \text{t} \hat{\mathbf{u}}_{\text{y}}\)

\(\therefore \quad r^{2}=(X-a \cos \omega t)^{2}+(Y-a \sin \omega t)^{2}+Z^{2}\)

\(r^{2}=X^{2}+Y^{2}+Z^{2}+a^{2}-2 a X \cos \omega t-2 a Y \sin \omega t\)

or \(r^{2}=R^{2}\left[1+\left(\frac{a}{R}\right)^{2}-\frac{2 a X}{R^{2}} \cos \omega t-\frac{2 a Y}{R^{2}} \sin \omega t\right]\).

Keep only the lowest terms in \(\left(\frac{a}{R}\right)\) :

\(r \cong R\left[1-\frac{2 a X}{R^{2}} \cos \omega t-\frac{2 a Y}{R^{2}} \sin \omega t\right]^{1 / 2}\)

So \(\frac{1}{r^{3}} \cong \frac{1}{R^{3}}\left[1-\frac{2 a X}{R^{2}} \cos \omega t-\frac{2 a Y}{R^{2}} \sin \omega t\right]^{-3 / 2}\)

\(\frac{1}{r^{3}} \cong \frac{1}{R^{3}}\left[1+\frac{3 a X}{R^{2}} \cos \omega t+\frac{3 a Y}{R^{2}} \sin \omega t\right]\)

\(\mathbf{E}_{\text{p}}=\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{\mathbf{r}}{r^{3}}\right)=\frac{q}{4 \pi \varepsilon_{0}} \frac{(\mathbf{R}-\mathbf{a})}{\text{R}^{3}}\left[1+\frac{3 a X}{R^{2}} \cos \omega t+\frac{3 a Y}{R^{2}} \sin \omega t\right]\)

Now multiply out and take time averages.

\(<\cos \omega t>=<\sin \omega t>=<\sin \omega t \quad \cos \omega t>=0\)

\(<\cos ^{2} \omega t>=<\sin ^{2} \omega t>=1 / 2\)

Notice that all terms proportional to a average to zero.

(a) \(\therefore \quad\left\langle\mathbf{E}_{\text{p}}\right\rangle \cong \frac{q}{4 \pi \varepsilon_{0}} \quad\left(\frac{\mathbf{R}}{R^{3}}\right)\). The correction terms are of order \(\left(\frac{a}{R}\right)^{2}\).

(b) \(\mathbf{B}_{\text{p}}=\frac{1}{\text{c}^{2}} \quad(\mathbf{V} \quad \times \mathbf{E})\)

\(\mathbf{v}=\frac{\mathbf{d a}}{d t}=-a \omega \sin \omega t \quad \hat{\mathbf{u}}_{x}+a \omega \cos \omega t \quad \hat{\mathbf{u}}_{y}\)

\(\therefore \quad \mathbf{B}_{p} \cong \frac{1}{c^{2}}\left(\frac{q}{4 \pi \varepsilon_{0}}\right) \frac{[(\mathbf{v} \times \mathbf{R})-(\mathbf{v} \times \mathbf{a})]}{\mathbb{R}^{3}}\left\{1+\frac{3 a X}{R^{2}} \cos \omega t\right. + \left.\frac{3 a Y}{R^{2}} \sin \omega t\right\}\)

\(\mathbf{v} \times \mathbf{R}=\left(\begin{array}{lll}
a \omega Z & \cos \omega t
\end{array}\right) \quad \hat{\mathbf{u}}_{x}+(a \omega z \sin \omega t) \hat{\mathbf{u}}_{y} - [\text{a} \omega \text{Y} \sin \omega \text{t}+\text{a} \omega \text{X} \cos \omega \text{t}] \hat{\mathbf{u}}_{\text{z}}\)

\(\mathbf{v} \times \mathbf{a}=-\text{a}^{2} \omega \hat{\mathbf{u}}_{z}\)

Multiply out the terms in B_p and take time averages. The result is

\(\left\langle\mathbf{B}_{\text{p}}\right\rangle=\frac{1}{\text{c}^{2}}\left(\frac{\text{q}}{4 \pi \varepsilon_{\text{O}}}\right) \frac{1}{\text{R}^{3}}\left\{\left(\frac{3 \text{a}^{2} \omega \text{XZ}}{2 \text{R}^{2}}\right) \hat{\mathbf{u}}_{\text{x}}+\left(\frac{3 \text{a}^{2} \omega \text{YZ}}{2 \text{R}^{2}}\right) \hat{\mathbf{u}}_{\text{y}}-\right. - \left.\left(\frac{3 a^{2} \omega}{2 R^{2}}\right) \quad\left(X^{2}+Y^{2}\right) \quad \mathbf{\hat{u}}_{z}+a^{2} \omega \mathbf{\hat{u}}_{z}\right\}\)

add and subtract \(\frac{3 a^{2} \omega}{2 R^{2}} z^{2} \mathbf{\hat{u}}_{z}\) to obtain

\(<\mathbf{B}_{p}>\frac{1}{c^{2}}\left(\frac{q}{4 \pi \varepsilon_{0}}\right) \frac{1}{R^{3}}\left\{\left(\frac{3 a^{2} \omega}{2 R^{2}}\right) \quad Z \mathbf{R}-\left(\frac{a^{2} \omega}{2}\right) \mathbf{\hat{u}}_{z}\right\}\).

Now \(\frac{1}{c^{2}}=\varepsilon_{0} \mu_{0}\) and \(\mathbf{m}=\frac{\text{qa}^{2} \omega}{2} \hat{\mathbf{u}}_{\text{z}}\)

Therefore \(\left\langle\mathbf{B}_{p}\right\rangle=\left(\frac{\mu_{0}}{4 \pi}\right)\left[\frac{3(\mathbf{m} \cdot \mathbf{R}) \mathbf{R}}{R^{5}}-\frac{\mathbf{m}}{R^{3}}\right]\).

Problem (1.9)

Given the following scalar functions, V, expressed in cylindrical polar co-ordinates. For each function calculate

(1) the components of grad V

(2) ∇² V

(a) V = r Cosθ

(b) V = ln r

(d) \(V=\frac{\cos n \theta}{r^{n}}\), where n is an integer either positive or negative.

Answer (1.9)

\(\operatorname{grad} V=\left(\frac{\partial V}{\partial r}\right) \mathbf{\hat{u}}_{r}+\frac{1}{r}\left(\frac{2 V}{\partial \theta}\right) \mathbf{\hat{u}}_{\theta}+\left(\frac{\partial V}{\partial z}\right) \mathbf{\hat{u}}_{z}\)

\(\nabla^{2} V=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial V}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial \theta^{2}}+\frac{\partial^{2} V}{\partial z^{2}}\)

(a) V= rCosθ.

\(\left.\operatorname{grad} V\right|_{r}=\cos \theta\)

\(\left.\operatorname{grad} V\right|_{\theta}=-\sin \theta\)

These correspond to a unit vector along x!

∇²V = 0

(b) V= lnr.

\(\left.\operatorname{grad} V\right|_{r}=\frac{1}{r}\)

\(\left.\operatorname{grad} V\right|_{\theta}=0\)

∇²V = 0

\(\left.\operatorname{grad} V\right|_{r}=-\frac{\cos \theta}{r^{2}}\)

\(\left.\operatorname{grad} V\right|_{\theta}=-\frac{\sin \theta}{r^{2}}\)

∇²V = 0

(d) \(\text{V}=\frac{\cos \text{n} \theta}{\text{r}^{\text{n}}}.\)

\(\left.\operatorname{gradv}\right|_{r}=-\frac{n \cos n \theta}{r^{n+1}}\)

\(\left.\operatorname{grad} V\right|_{\theta}=-\frac{\operatorname{nsin}(n \theta)}{r^{n+1}}\)

∇²V = 0 for any n.

Problem (1.10)

Given the following scalar functions V expressed in spherical polar co-ordinates. For each function calculate

(1) the components of grad V

(2) ∇²V

(a) V = r Cosθ

(b) \(V=\frac{\operatorname{Cos} \theta}{r^{2}}\)

(d) \(V=\frac{\left(3 \cos ^{2} \theta-1\right)}{r^{3}}\)

(e) \(V=\frac{\cos n \theta}{r^{n}}\) where n is a positive integer.

Answer (1.10)

\(\nabla \text{V} \quad=\left(\frac{\partial \text{V}}{\partial \text{r}}\right) \hat{\mathbf{u}}_{\text{r}}+\frac{1}{\text{r}}\left(\frac{\partial \text{V}}{\partial \theta}\right) \hat{\mathbf{u}}_{\theta}+\frac{1}{\text{r} \sin \theta}\left(\frac{\partial \text{V}}{\partial \phi}\right) \hat{\mathbf{u}}_{\phi}\)

\(\nabla^{2} V=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial V}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial V}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2} V}{\partial \phi^{2}}\)

(a) V = rCosθ

\(\left.\operatorname{grad} V\right|_{r}=\cos \theta\)

\(\left.\operatorname{gradv}\right|_{\theta}=-\sin \theta\)