91Ó°ÊÓ

The given equation 11.60

${\mathrm{P}}_{\mathrm{rad}}\left({\mathrm{t}}_0\right)â‰\dots \frac{{\mathrm{μ}}_0}{6\mathrm{Ï€³¦}}{\left[\stackrel{¨}{\mathrm{p}}\left({\mathrm{t}}_0\right)\right]}^2$

The expression for the pointing vector is

$\mathbf{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)}$ ………. (1)

Here, ${\mathbf{μ}}_{\mathbf{0}}$is the permeability of free space, $\mathbf{E}$is the electric field, $\mathbf{B}$is the magnetic field.

The retarded vector potential is

role="math" localid="1658736946261" $\mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{Ï€}}\mathbf{∫}\frac{\mathbf{J}\left(t-\frac{r}{c}\right)}{\mathbf{r}}\mathbf{»åÏ„}\mathbf{\text{'}}$ ………. (2)

Here,$\mathbf{J}$is the current density, $\mathbf{C}$is the speed of the light, $\mathbf{t}$ is the time, and role="math" localid="1658737028413" $\mathbf{r}$is the position vector.

The change in current is

$\mathbf{dI}\mathbf{\text{'}}\mathbf{=}\mathbf{²ú»åÏ•}\mathbf{\text{'}}\widehat{\mathbf{Ï•}}$ ………. (3)

Here, $\mathbf{b}$is the radius of the wire loop, and $\widehat{\mathbf{Ï•}}$ is the direction vector.

The maxwell equation for the electric field is

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{∇}\mathbf{V}\mathbf{-}\frac{\mathbf{∂}\mathbf{A}}{\mathbf{∂}\mathbf{t}}$ ………. (4)

Here, $\mathbf{E}$ is the electric field, role="math" localid="1658737189047" $\mathbf{V}$is the voltage.

The maxwell equation for the magnetic field is

$\mathbf{B}\mathbf{=}\mathbf{∇}\mathbf{×}\mathbf{A}$ ………. (5)

Here, $\mathbf{∇}\mathbf{×}\mathbf{A}$ is the curl of role="math" localid="1658737263083" $\mathbf{A}$.

The total radiated power is

role="math" localid="1658737288650" $\mathbf{P}\mathbf{=}\mathbf{∫}\mathbf{S}\mathbf{.}\mathbf{da}$ ………. (6)

Here,role="math" localid="1658737336377" $\mathbf{S}$is the pointing vector.

The magnetic dipole moment is

$\mathbf{μ}\mathbf{=}\mathbf{IA}$ ………. (7)

Here, $\mathbf{I}$ is the current and$\mathbf{A}$ is the area of the ring.

From the equation (2)

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫\frac{\mathrm{J}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\mathrm{»åÏ„}\text{'}$

Rearrange the equation,

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫\frac{\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}(\mathrm{J»åÏ„}\text{'})$

As we know,

$\mathrm{J»åÏ„}\text{'}=\mathrm{IdI}\text{'}$

Substitute $\mathrm{IdI}\text{'}$ for $\mathrm{J»åÏ„}\text{'}$in the above equation.

$A(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}(\mathrm{IdI}\text{'})$ ………. (8)

Here,

$\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}}\left(1+\frac{\mathrm{b}}{\mathrm{r}}\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\right)$

Now substitute $(-\mathrm{²õ¾\pm ²ÔÏ•}\text{'}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õÏ•}\text{'}\widehat{\mathrm{y}})$ for $\mathrm{»åÏ•}\text{'}$ in the equation (3)

$\mathrm{dI}\text{'}=\mathrm{b}(-\mathrm{²õ¾\pm ²ÔÏ•}\text{'}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õÏ•}\text{'}\widehat{\mathrm{y}})\widehat{\mathrm{f}}$

Also,

$\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)=\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}+\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\right)$

Substitute ${\mathrm{t}}_0$for $\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}$ in the above equation.

$\begin{array}{c}\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)=\mathrm{I}\left({\mathrm{t}}_0+\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\right)\\ =\mathrm{I}\left({\mathrm{t}}_0\right)+\stackrel{Ë™}{\mathrm{I}}\left({\mathrm{t}}_0\right)\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\\ =\mathrm{I}+\stackrel{Ë™}{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\end{array}$

Substitute $\mathrm{I}+\stackrel{Ë™}{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}$for $\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)$ in the equation (8)

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}âˆ\circledR \frac{1}{\mathrm{r}}\left(1+\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õ´Ú}\text{'}\right)\left((\mathrm{I}+\mathrm{I}\text{'}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'})\right)\mathrm{b}(-\mathrm{²õ¾\pm ²ÔÏ•}\text{'}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õÏ•}\text{'}\widehat{\mathrm{y}})\mathrm{»åÏ•}\text{'}\\ =\frac{{\mathrm{μ}}_0\mathrm{b}}{4\mathrm{Ï€°ù}}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\left(\left(\mathrm{I}+\mathrm{I}\text{'}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õ´Ú}\text{'}\right)+\mathrm{I}\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\text{'}\right)(-\mathrm{²õ¾\pm ²ÔÏ•}\text{'}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õÏ•}\text{'}\widehat{\mathrm{y}})\mathrm{»åÏ•}\text{'}\end{array}$

From the trigonometry,

$\begin{array}{c}\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\mathrm{²õ¾\pm ²ÔÏ•}\text{'}\mathrm{»åÏ•}\text{'}=\underset{0}{\overset{2\mathrm{Ï€}}{∫}}\mathrm{³¦´Ç²õÏ•}\text{'}\mathrm{»åÏ•}\text{'}=0\\ \underset{0}{\overset{2\mathrm{Ï€}}{∫}}\mathrm{²õ¾\pm ²ÔÏ•³¦´Ç²õÏ•}\text{'}\mathrm{»åÏ•}\text{'}=0\\ \underset{0}{\overset{2\mathrm{Ï€}}{∫}}{\cos }^2\mathrm{Ï•}\text{'}\mathrm{»åÏ•}\text{'}=\mathrm{Ï€}\end{array}$

Therefore, the equation (9) becomes

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0\mathrm{b}}{4\mathrm{Ï€°ù}}\mathrm{Ï€}\widehat{\mathrm{y}}\left[\stackrel{Ë™}{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ôθ}+\mathrm{I}\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{²õ¾\pm ²Ôθ}\right]\\ \mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4{\mathrm{r}}^2}\left[\stackrel{Ë™}{\mathrm{I}}\left(\frac{\mathrm{r}}{\mathrm{c}}\right)\mathrm{²õ¾\pm ²Ôθ}+\mathrm{I²õ¾\pm ²Ôθ}\right]\widehat{\mathrm{y}}\end{array}$

For points not lying (z-x) plane

$\widehat{y}→\widehat{f}$

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\left[\stackrel{Ë™}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{f}}$ ………. (10)

From the equation (4)

$\mathrm{E}(\mathrm{r},\mathrm{t})=-∇\mathrm{V}-\frac{∂\mathrm{A}}{∂\mathrm{t}}$

As the voltage, $\mathrm{V}=0$, the electric field is

$\mathrm{E}(\mathrm{r},\mathrm{t})=-\frac{∂\mathrm{A}}{∂\mathrm{t}}$

Substitute equation (10) in the above expression.

$\begin{array}{c}\mathrm{E}(\mathrm{r},\mathrm{t})=\frac{∂}{∂\mathrm{t}}\left(\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\left[\stackrel{Ë™}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{Ï•}}\right)\\ \mathrm{E}(\mathrm{r},\mathrm{t})=\frac{-{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\frac{∂}{∂\mathrm{t}}\stackrel{Ë™}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{Ï•}}\\ \mathrm{E}(\mathrm{r},\mathrm{t})=\frac{-{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\stackrel{¨}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{Ï•}}\end{array}$

From the equation (5)

$\mathrm{B}=∇×\mathrm{A}$

The curl of the A is

$∇×\mathrm{A}=\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂}{∂\mathrm{θ}}\left({\mathrm{A}}_{\mathrm{Ï•}}\mathrm{²õ¾\pm ²Ôθ}\right)\widehat{\mathrm{r}}-\frac{1}{\mathrm{r}}\frac{∂}{∂\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{Ï•}}\right)\widehat{\mathrm{θ}}$

Substitute equation (10) in the above expression

Therefore, the magnetic field is$\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\stackrel{¨}{\mathrm{I}}\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{θ}}$

From the equation (1)

$\mathrm{S}=\frac{1}{{\mathrm{μ}}_0}(\mathrm{E}×\mathrm{B})$

Substitute$\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\stackrel{¨}{\mathrm{I}}\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{θ}}$ for $\mathrm{B}$ and $\frac{-{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\stackrel{¨}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{Ï•}}$ for $\mathrm{E}$

$\begin{array}{c}\mathrm{S}=\frac{1}{{\mathrm{μ}}_0}\left(\frac{-{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\stackrel{¨}{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{f}}2×\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\stackrel{¨}{\mathrm{I}}\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\widehat{\mathrm{θ}}\right)\\ \mathrm{S}=\frac{1}{{\mathrm{μ}}_0\mathrm{c}}{\left(\frac{{\mathrm{μ}}_0{\mathrm{b}}^2}{4\mathrm{c}}\stackrel{¨}{\mathrm{I}}\frac{\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{r}}\right)}^2(-\widehat{\mathrm{f}}×\widehat{\mathrm{θ}})\\ \mathrm{S}=\frac{{\mathrm{μ}}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{θ}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}\end{array}$

The total radiated power is

$\mathrm{P}=∫\mathrm{S}.\mathrm{da}$

Substitute$\frac{{\mathrm{μ}}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{θ}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$ for$\mathrm{S}$.

$\begin{array}{c}\mathrm{P}=∫\mathrm{S}.\mathrm{da}\\ \mathrm{P}=\underset{\mathrm{f}=0}{\overset{2\mathrm{Ï€}}{∫}}\underset{\mathrm{θ}=0}{\overset{\mathrm{Ï€}}{∫}}\frac{{\mathrm{μ}}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{θ}}{{\mathrm{r}}^2}{\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθ»åθdf}\\ \mathrm{P}=2\mathrm{Ï€}\frac{{\mathrm{μ}}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2\underset{0}{\overset{\mathrm{Ï€}}{∫}}{\sin }^2\mathrm{θ»åθ}\\ \mathrm{P}=\frac{{\mathrm{μ}}_0}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2×2\mathrm{Ï€}×\frac{4}{3}\end{array}$

Hence,

$\mathrm{P}=\frac{{\mathrm{μ}}_0\mathrm{π}}{6{\mathrm{c}}^3}{\left({\mathrm{b}}^2\stackrel{¨}{\mathrm{I}}\right)}^2$ ………… (11)

From the equation (7)

$\mathrm{μ}=\mathrm{IA}$

Substitute ${\mathrm{Ï€²ú}}^2$ for $\mathrm{A}$

$\mathrm{μ}=\mathrm{I}×\left({\mathrm{Ï€²ú}}^2\right)$

Here, $I$is the current and $b$is the radius of the ring.

Differentiate the above equation twice.

$\begin{array}{c}\stackrel{¨}{\mathrm{m}}=\stackrel{¨}{\mathrm{I}}{\mathrm{Ï€²ú}}^2\\ \frac{\stackrel{¨}{\mathrm{m}}}{\mathrm{Ï€}}=\stackrel{¨}{\mathrm{I}}{\mathrm{b}}^2\end{array}$

Substitute $\frac{\stackrel{¨}{\mathrm{m}}}{\mathrm{π}}$ for $\stackrel{¨}{\mathrm{I}}{\mathrm{b}}^2$in the equation (11)

$\begin{array}{c}\mathrm{P}=\frac{{\mathrm{μ}}_0\mathrm{Ï€}}{6{\mathrm{c}}^3}{\left(\frac{\stackrel{¨}{\mathrm{m}}}{\mathrm{Ï€}}\right)}^2\\ \mathrm{P}=\frac{{\mathrm{μ}}_0\mathrm{Ï€}{\stackrel{¨}{\mathrm{m}}}^2}{6{\mathrm{c}}^3×{\mathrm{Ï€}}^2}\\ \mathrm{P}=\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{m}}}^2}{6{\mathrm{Ï€³¦}}^3}\end{array}$

Thus, the general formula for the power radiated is $\mathrm{P}=\frac{{\mathrm{μ}}_0{\stackrel{¨}{\mathrm{m}}}^2}{6{\mathrm{Ï€³¦}}^3}$.