Obtain the earth's magnetization, assuming that the earth's field can be approximated by a giant bar magnet of magnetic moment $8.0 \times 10^{22} A m^{2}$ . Radius of earth = $6400 k m$ .

62.8 A m^{- 1}

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52.8 A m^{- 1}

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72.8 A m^{- 1}

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82.8 A m^{- 1}

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Solution

The correct option is C $72.8 A m^{- 1}$
Given,
Magnetic moment, $M = 8.0 \times 10^{22} A m^{2}$
$R = 64000$ km
$= 64 \times 10^{5}$ m
Magnetization of earth is given by,
$\begin{matrix} I = \frac{M}{V} = \frac{M}{\frac{4}{3} π R^{3}} = \frac{3 M}{4 π R^{3}} = \frac{3 \times 8.0 \times 10^{22}}{4 π \times {(64 \times 10^{5})}^{3}} = 72.8 A m^{- 1} \end{matrix}$
Hence, Option $C$ is correct.

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Obtain the earth's magnetization, assuming that the earth's field can be approximated by a giant bar magnet of magnetic moment 8.0×1022Am2. Radius of earth = 6400km.

The correct option is C 72.8 Am−1Given,Magnetic moment, M=8.0×1022Am2R=64000 km=64×105 mMagnetization of earth is given by,I=MV=M43πR3=3M4πR3=3×8.0×10224π×(64×105)3=72.8Am−1Hence, Option C is correct.

Obtain the earth's magnetization, assuming that the earth's field can be approximated by a giant bar magnet of magnetic moment $8.0 \times 10^{22} A m^{2}$ . Radius of earth = $6400 k m$ .