JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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 \({10^{10}}\,A\,{m^2}\)
Answer & Solution
Correct Answer
(A) \({10^{23}}\,A\,{m^2}\)
Step-by-step Solution
Detailed explanation
Given, \(B=4 \times 10^{-5} \,\mathrm{T}\) \(\mathrm{R}_{\mathrm{E}}=6.4 \times 10^{6}\, \mathrm{m}\) Dipole moment of the earth \(\mathrm{M}=?\) \(B = \frac{{{\mu _0}M}}{{4\pi {d^3}}}\)…
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