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Question

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{5} \ G$

Vedclass · Answer

(A) The angle of dip $\delta$ is the angle that the total magnetic field of the earth makes with the horizontal direction. Here,$\delta = 30^{\circ}$.Given,the horizontal component of the earth's magnetic field is $B_H = 0.3 \ G$.The relationship between the total magnetic field $B$,the horizontal component $B_H$,and the angle of dip $\delta$ is given by $B_H = B \cos \delta$.Substituting the given values: $0.3 = B \cos 30^{\circ}$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $0.3 = B 	imes \frac{\sqrt{3}}{2}$.Solving for $B$: $B = \frac{0.3 	imes 2}{\sqrt{3}} = \frac{0.6}{\sqrt{3}} = \frac{0.6 	imes \sqrt{3}}{3} = 0.2 \sqrt{3} \ G$.Alternatively,$B = \frac{0.6}{\sqrt{3}} = \frac{6}{10 \sqrt{3}} = \frac{3}{5 \sqrt{3}} = \frac{\sqrt{3}}{5} \ G$.

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