At a certain place,the angle of dip is 60^{ci | vedclass

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(D) Given:Horizontal component of earth's magnetic field,$B_H = 0.8 	imes 10^{-4} ~T$Angle of dip,$	heta = 60^{\circ}$We need to find the earth's to

Vedclass · Accepted Answer

$1.6 	imes 10^{-4} ~T$

Vedclass · Answer

(D) Given:Horizontal component of earth's magnetic field,$B_H = 0.8 	imes 10^{-4} ~T$Angle of dip,$	heta = 60^{\circ}$We need to find the earth's total magnetic field,$B_e$.The relationship between the total magnetic field,its horizontal component,and the angle of dip is given by:$B_H = B_e \cos 	heta$Substituting the given values:$0.8 	imes 10^{-4} = B_e \cos 60^{\circ}$Since $\cos 60^{\circ} = 0.5$:$0.8 	imes 10^{-4} = B_e 	imes 0.5$Solving for $B_e$:$B_e = \frac{0.8 	imes 10^{-4}}{0.5}$$B_e = 1.6 	imes 10^{-4} ~T$Thus,the earth's overall magnetic field is $1.6 	imes 10^{-4} ~T$.

At a certain place,the angle of dip is $60^{\circ}$ and the horizontal component of the earth's magnetic field $(B_H)$ is $0.8 \times 10^{-4} ~T$. The earth's overall magnetic field is

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