As you know, the magnetic north pole of the earth does not coincide with the geographic north pole—in fact, it’s off by about $\mathbf{11}\mathbf{°}$. Relative to the fixed axis of rotation, therefore, the magnetic dipole moment of the earth is changing with time, and the earth must be giving off magnetic dipole radiation.

(a) Find the formula for the total power radiated, in terms of the following parameters:$\mathit{\psi }$ (the angle between the geographic and magnetic north poles), M (the magnitude of the earth’s magnetic dipole moment), and$\mathit{\omega }$ (the angular velocity of rotation of the earth). [Hint: refer to Prob. 11.4 or Prob. 11.11.]

(b) Using the fact that the earth’s magnetic field is about half a gauss at the equator, estimate the magnetic dipole moment Mof the earth.

(c) Find the power radiated. [Answer: $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{W}$]

(d) Pulsars are thought to be rotating neutron stars, with a typical radius of 10 km, a rotational period of ${\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{s}$, and a surface magnetic field of ${\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{T}$. What sort of radiated power would you expect from such a star? [Answer: $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{36}}\mathbf{}\mathbf{W}$].

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

$\mathit{P}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{m}}_{\mathbf{0}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{4}}}{\mathbf{12}\mathbf{\Pi }{\mathbf{c}}^{\mathbf{3}}}$ …… (1)

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

$m_0=M\sin \psi$

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

(a)

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

$m\left(t\right)=M\cos \psi \hat{z}+M\sin \psi \left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right]$

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.

$$\begin{gathered} P=2\left(\frac{{\mu }_0{m}_0^2{\omega }^4}{12\pi c^3}\right) \\ P=\frac{{\mu }_0{\left(M\sin \psi \right)}^2{\omega }^4}{6\pi c^3}........\left(2\right) \end{gathered}$$

Therefore, the formula for the total power radiated in terms of$\psi ,M$ and$\omega$ is

$P=\frac{{\mu }_0{\left(M\sin \psi \right)}^2{\omega }^4}{6\pi c^3}$.

(b)

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

$B=\frac{{\mu }_{0}m}{4\pi r^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)$

Here,$\theta =\frac{\pi }{2}$and$r=R$and $m=M$.

Hence, the above expression becomes,

$$\begin{gathered} B=\frac{{\mu }_{0}M}{4\pi R^3}\left(2\cos \left(\frac{\pi }{2}\right)\hat{r}+\sin \left(\frac{\pi }{2}\right)\hat{\theta }\right) \\ B=\frac{{\mu }_{0}M}{4\pi R^3} \\ M=\frac{B\left(4\pi R^3\right)}{{\mu }_0} \end{gathered}$$

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

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

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

(c)

Substitute${\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m},M=1.31\times {10}^{23}\mathrm{A}·{\mathrm{m}}^2,\varphi =11°,\omega =\frac{2\pi }{T}$ and$c=3\times {10}^8\mathrm{m}/\mathrm{s}$ in equation (2).

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

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

(d)

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

$$\begin{gathered} P=\frac{{\mu }_0{\left({\displaystyle \frac{B\left(4\pi R^3\right)}{{\mu }_0}}\sin \psi \right)}^2{\omega }^4}{6\pi c^3} \\ P=\frac{16{\pi }^2{R}^6{B}^2}{{\mu }_{0}6\pi c^3}{\omega }^4{\sin }^2\psi \end{gathered}$$

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

$$\begin{gathered} P=\frac{16{\pi }^2{R}^6{B}^2}{{\mu }_{0}6\pi c^3}{\omega }^4\left(\frac{1}{2}\right) \\ P=\frac{4{\pi }^2{R}^2{B}^2}{3{\mu }_0\pi c^3}{\omega }^4 \end{gathered}$$

Substitute$R=10km,B=10T,{\mu }_0=4\pi \times {10}^{-7}H/m,c=3\times {10}^{8}m/s,\omega =\frac{2\pi }{T}$ and$T={10}^{-3}S$ in the above expression.

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

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