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5.Magnetism and Matter
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The magnetic field of earth at the equator is approximately $4 \times 10^{-5}\, T$ . The radius of earth is $6.4 \times 10^6\, m$. Then the dipole moment of the earth will be nearly of the order of
A
${10^{23}}\,A\,{m^2}$
B
${10^{20}}\,A\,{m^2}$
C
${10^{16}}\,A\,{m^2}$
D
${10^{10}}\,A\,{m^2}$
(JEE MAIN-2014)
Solution
Given, $B=4 \times 10^{-5} \,\mathrm{T}$
$\mathrm{R}_{\mathrm{E}}=6.4 \times 10^{6}\, \mathrm{m}$
Dipole moment of the earth $\mathrm{M}=?$
$B = \frac{{{\mu _0}M}}{{4\pi {d^3}}}$
$4 \times {10^{ – 5}} = \frac{{4\pi \times {{10}^{ – 7}} \times {\text{M}}}}{{4\pi \times \left( {6.4 \times {{10}^6}} \right.{)^3}}}$
$\mathrm{M} \cong 10^{23}\, \mathrm{Am}^{2}$
Standard 12
Physics
