Question

# The magnitude of the magnetic dipole moment of Earth is $8.0 \times 10^{22} J/T$. If the origin of this magnetism were a magnetized iron sphere at the center of Earth, what would be its radius?

Solution

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(a) The total dipole moment of a saturated sphere with iron atoms is given by:

\begin{aligned} \mu_{tot.}=N\mu \end{aligned}

where $N$ is the number of iron atoms in the sphere. The mass the of an iron sphere can be written as:

\begin{aligned} M=Nm \end{aligned}

where $m$ is the mass of a single atom, and also it can be written in terms of the density and the volume of the sphere:

$M=\rho V$

\begin{aligned} M=\dfrac{4}{3} \pi \rho R^3 \end{aligned}

where $V=4/3 \pi R^3$, by equating (2) and (3) we get:

$Nm=\dfrac{4}{3}\rho \pi R^3$

$N=\dfrac{4\pi \rho R^3}{3m}$

substitute into (1) to get:

$\mu_{tot.}=\dfrac{4\pi \rho R^3 \mu}{3m}$

solve for $R$ to get:

\begin{aligned} R= \left(\dfrac{3m \mu_{tot.}}{4\pi\rho \mu} \right)^{1/3} \end{aligned}

the mass of the iron atom is:

$m=56 \mathrm{~u} \times \dfrac{1.66 \times 10^{-27} \mathrm{~kg}}{1\mathrm{~u}}=9.30 \times 10^{-26} \mathrm{~kg}$

substitute with the givens into (4) to get:

\begin{aligned} R&= \left(\dfrac{3(9.30 \times 10^{-26} \mathrm{~kg}) (8.0 \times 10^{22} \mathrm{~J/T})}{4\pi(14 \times 10^{3} \mathrm{~kg/m^3}) (2.1 \times 10^{-23} \mathrm{~J/T})} \right)^{1/3} \\ &=1.82 \times 10^{5} \mathrm{~m} \end{aligned}

$\boxed{R=1.82 \times 10^{5} \mathrm{~m}}$

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