The horizontal component of the Earth's magne | vedclass

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(C) Given:Horizontal component of Earth's magnetic field,$B_H = 0.22 \ G$Total magnetic field,$B = 0.4 \ G$Let $\phi$ be the angle of dip.The horizont

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$	an^{-1}(1.518)$

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(C) Given:Horizontal component of Earth's magnetic field,$B_H = 0.22 \ G$Total magnetic field,$B = 0.4 \ G$Let $\phi$ be the angle of dip.The horizontal component is given by $B_H = B \cos \phi$.Therefore,$\cos \phi = \frac{B_H}{B} = \frac{0.22}{0.4} = 0.55$.The vertical component is $B_V = B \sin \phi = \sqrt{B^2 - B_H^2} = \sqrt{(0.4)^2 - (0.22)^2} = \sqrt{0.16 - 0.0484} = \sqrt{0.1116} \approx 0.334 \ G$.The angle of dip $\phi$ is given by $	an \phi = \frac{B_V}{B_H} = \frac{\sqrt{B^2 - B_H^2}}{B_H} = \frac{\sqrt{(0.4)^2 - (0.22)^2}}{0.22} = \frac{\sqrt{0.1116}}{0.22} \approx \frac{0.334}{0.22} \approx 1.518$.Thus,$\phi = 	an^{-1}(1.518)$.

The horizontal component of the Earth's magnetic field is $0.22 \ G$ and the total magnetic field is $0.4 \ G$. The angle of dip is:

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