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(D) Given,magnetic moment $M = 8 	imes 10^{22} 	ext{ Am}^2$.Radius of the earth $R_e = 6.4 	imes 10^6 	ext{ m}$.The magnetic field $B$ at the equa

Vedclass · Accepted Answer

$0.32$

Vedclass · Answer

(D) Given,magnetic moment $M = 8 	imes 10^{22} 	ext{ Am}^2$.Radius of the earth $R_e = 6.4 	imes 10^6 	ext{ m}$.The magnetic field $B$ at the equator for a magnetic dipole is given by the formula:$B = \frac{\mu_0}{4\pi} \cdot \frac{M}{R_e^3}$Substituting the values:$B = 10^{-7} 	imes \frac{8 	imes 10^{22}}{(6.4 	imes 10^6)^3}$$B = \frac{8 	imes 10^{15}}{262.144 	imes 10^{18}}$$B \approx 0.0305 	imes 10^{-3} 	ext{ T}$Since $1 	ext{ T} = 10^4 	ext{ Gauss}$,$B \approx 0.0305 	imes 10^{-3} 	imes 10^4 	ext{ Gauss} = 0.305 	ext{ Gauss}$.Rounding to the nearest option,the value is approximately $0.32 	ext{ Gauss}$.

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