In what way is Gauss’s law in magnetism different from that used in electrostatics? Explain briefly. The Earth’s magnetic field at the Equator is approximately 0.4 G. Estimate the Earth’s magnetic dipole moment. Given: Radius of the Earth = 6400 km.
CBSE OUTSIDE DELHI - SET 2 PATNA 2015
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In magnetism, Gauss’s law states: $\oint\overrightarrow{\text{B}}.\overrightarrow{\text{ds}} = 0 $
In electrostatistics, Gauss’s law states: $\oint\overrightarrow{\text{B}}.\overrightarrow{\text{ds}} = \frac{\text{q}}{\varepsilon_\circ}$
Reason: Isolated magnetic poles do not exist,
$\text{B} = \frac{\mu_{\circ}}{4\pi}\bigg(\frac{\text{m}}{\text{R}^{3}}\bigg) = 10^{-7}\bigg(\frac{\text{m}}{\text{R}^{3}}\bigg)$
$\text{m} = \frac{0.4\times 10^{−4}\times (6400 \times 10^{3})^{3}}{10^{-7}}$
$ = 1.1\times10^{23}\text{Am}^{2}.$
In electrostatistics, Gauss’s law states: $\oint\overrightarrow{\text{B}}.\overrightarrow{\text{ds}} = \frac{\text{q}}{\varepsilon_\circ}$
Reason: Isolated magnetic poles do not exist,
$\text{B} = \frac{\mu_{\circ}}{4\pi}\bigg(\frac{\text{m}}{\text{R}^{3}}\bigg) = 10^{-7}\bigg(\frac{\text{m}}{\text{R}^{3}}\bigg)$
$\text{m} = \frac{0.4\times 10^{−4}\times (6400 \times 10^{3})^{3}}{10^{-7}}$
$ = 1.1\times10^{23}\text{Am}^{2}.$
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