Question

The earth's magnetic field at some place on magnetic equator of the earth is $0.5\times {10}^{-4}T$. Consider the radius of the earth at that place as $6400km$. Then, magnetic dipole moment of the earth is ___________ ${Am}^2$. $({\mu }_0=4\pi \times {10}^{-7}Tm{A}^{-1})$

• A. $1.31\times {10}^{23}$
• B. $1.05\times {10}^{23}$
• C. $1.15\times {10}^{23}$
• D. $1.62\times {10}^{23}$

Answer 1

The correct option is D $1.62\times {10}^{23}$

Given,
$R_e=6400km=6.4\times {10}^{+6}m$
$B_e=0.5\times {10}^{-4}T$
${\mu }_0=4\pi \times {10}^{-7}Tm{A}^{-1}$
$M=$?
The magnetic field at the equator
$B=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{R_e}$
and $M=iA$
$M=\frac{4\pi R_e\cdot B}{{\mu }_0}\times \pi R_e^2$
$=\frac{4\pi \times 64\times {10}^5\times 5\times {10}^{-5}\times 3.14\times 4096\times {10}^{10}}{4\pi \times {10}^{-7}}$
$=130\times {10}^7\times 3.14\times 4096\times {10}^{10}$
$=1673508.57\times {10}^{17}$
$=1.67\times {10}^{23}{Am}^2$