91Ó°ÊÓ

Write the expression for the total power radiated by the dipole of dipole moment.

$\mathit{P}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{m}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{Ó\neg }}^{\mathbf{4}}}{\mathbf{12}\mathbf{Î }{\mathbf{c}}^{\mathbf{3}}}$ …… (1)

Here,${\mathrm{μ}}_0$ is the permeability of the free space,$\mathrm{Ó\neg }$ is the angular velocity, c is the speed of light and$m_0$ is the magnetic dipole moment which is given as:

$m_0=M\sin \mathrm{ψ}$

Here, M is the magnitude of the earth’s magnetic dipole moment and$\mathrm{ψ}$ is the angle between the geographic and magnetic north poles.

(a)

Write the expression for the magnetic dipole moment vector of the earth.

$m\left(t\right)=M\cos \mathrm{ψ}\hat{z}+M\sin \mathrm{ψ}\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{x}+\sin \left(\mathrm{Ó\neg }t\right)\hat{y}\right]$

Consider the relation between the power radiated by the magnetic dipole moment and the power radiated by oscillating magnetic dipole of dipole moment.

$P=2P\text{'}$

Substitute the value of equation (1) in the above expression.

$$\begin{gathered} P=2\left(\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{12\mathrm{Ï€}{c}^3}\right) \\ P=\frac{{\mathrm{μ}}_0{\left(M\sin \mathrm{ψ}\right)}^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}{c}^3}........\left(2\right) \end{gathered}$$

Therefore, the formula for the total power radiated in terms of$\mathrm{ψ},M$ and$\mathrm{Ó\neg }$ is

$P=\frac{{\mathrm{μ}}_0{\left(M\sin \mathrm{ψ}\right)}^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}{c}^3}$.

(b)

Write the expression for the magnetic field due to a magnetic dipole.

$B=\frac{{\mathrm{μ}}_{0}m}{4\mathrm{π}{r}^3}\left(2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)$

Here,$\mathrm{θ}=\frac{\mathrm{π}}{2}$and$r=R$and $m=M$.

Hence, the above expression becomes,

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}M}{4\mathrm{π}{R}^3}\left(2\cos \left(\frac{\mathrm{π}}{2}\right)\hat{r}+\sin \left(\frac{\mathrm{π}}{2}\right)\widehat{\mathrm{θ}}\right) \\ B=\frac{{\mathrm{μ}}_{0}M}{4\mathrm{π}{R}^3} \\ M=\frac{B\left(4\mathrm{π}{R}^3\right)}{{\mathrm{μ}}_0} \end{gathered}$$

Substitute $B=0.5\mathrm{gauss},R=6.4×{10}^6\mathrm{m}$and${\mathrm{μ}}_0=4\mathrm{π}×{10}^{-7}\mathrm{H}/\mathrm{m}$ in the above expression.

$$\begin{gathered} M=\frac{\left(0.5\mathrm{gauss}×{\displaystyle \frac{{10}^{-4}\mathrm{T}}{1\mathrm{gauss}}}\right)\left(4\mathrm{π}{\left(6.4×{10}^5\mathrm{m}\right)}^3\right)}{\left(4\mathrm{π}×{10}^{-7}\mathrm{H}/\mathrm{m}\right)} \\ M=1.31×{10}^{23}\mathrm{A}·{\mathrm{m}}^2 \end{gathered}$$

Therefore, the magnetic dipole moment M of the earth is $1.31×{10}^{23}\mathrm{A}·{\mathrm{m}}^2$.

(c)

Substitute${\mathrm{μ}}_0=4\mathrm{Ï€}×{10}^{-7}\mathrm{H}/\mathrm{m},M=1.31×{10}^{23}\mathrm{A}·{\mathrm{m}}^2,\mathrm{Ï•}=11°,\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$ and$c=3×{10}^8\mathrm{m}/\mathrm{s}$ in equation (2).

$$\begin{gathered} P=\frac{\left(4\mathrm{π}×{10}^{-7}\mathrm{H}/\mathrm{m}\right){\left(1.31×{10}^{23}\mathrm{A}·{\mathrm{m}}^2\sin 11°\right)}^2{\left({\displaystyle \frac{2\mathrm{π}}{24×60×60}}\mathrm{rad}/\mathrm{s}\right)}^4}{6\mathrm{π}{\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^3} \\ P=\frac{\left(2.499×{10}^{38}\right)×\left(2.796×{10}^{-17}\right)}{1.62×{10}^{26}} \\ P=4.31×{10}^{-5}W≈4×{10}^{-5}W \\ P=4×{10}^{-5}W \end{gathered}$$

Therefore, the power radiated is$4×{10}^{-5}W$ .

(d)

Substitute $M=\frac{B\left(4\mathrm{π}{R}^3\right)}{{\mathrm{μ}}_0}$in equation (2).

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{\left({\displaystyle \frac{B\left(4\mathrm{Ï€}{R}^3\right)}{{\mathrm{μ}}_0}}\sin \mathrm{ψ}\right)}^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}{c}^3} \\ P=\frac{16{\mathrm{Ï€}}^2{R}^6{B}^2}{{\mathrm{μ}}_{0}6\mathrm{Ï€}{c}^3}{\mathrm{Ó\neg }}^4{\sin }^2\mathrm{ψ} \end{gathered}$$

Use the average value$\frac{1}{2}$ for ${\sin }^2\mathrm{ψ}$. Hence,

$$\begin{gathered} P=\frac{16{\mathrm{Ï€}}^2{R}^6{B}^2}{{\mathrm{μ}}_{0}6\mathrm{Ï€}{c}^3}{\mathrm{Ó\neg }}^4\left(\frac{1}{2}\right) \\ P=\frac{4{\mathrm{Ï€}}^2{R}^2{B}^2}{3{\mathrm{μ}}_0\mathrm{Ï€}{c}^3}{\mathrm{Ó\neg }}^4 \end{gathered}$$

Substitute$R=10km,B=10T,{\mathrm{μ}}_0=4\mathrm{Ï€}×{10}^{-7}H/m,c=3×{10}^{8}m/s,\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$ and$T={10}^{-3}S$ in the above expression.

$$\begin{gathered} P=\frac{4{\mathrm{π}}^2{\left(10km×{\displaystyle \frac{{10}^{3}m}{1km}}\right)}^6{\left({10}^{8}T\right)}^2}{3\left(4\mathrm{π}×{10}^{-7}H/m\right)\mathrm{π}{\left(3×{10}^{8}m/s\right)}^3}{\left(\frac{2\mathrm{π}}{{10}^{-3}s}\right)}^4 \\ P=1.91×{10}^{36}W≈2×{10}^{36}W \\ P=2×{10}^{36}W \end{gathered}$$

Therefore, the radiated power that would be expected from a star is$2×{10}^{36}W$ .