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(A) Given: Horizontal component of Earth's magnetic field,$B_H = 0.3 \; Oersted$.Total intensity of Earth's magnetic field,$I = 0.6 \; Oersted$.The re

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(A) Given: Horizontal component of Earth's magnetic field,$B_H = 0.3 \; Oersted$.Total intensity of Earth's magnetic field,$I = 0.6 \; Oersted$.The relationship between the horizontal component,total intensity,and the angle of dip $(\phi)$ is given by: $B_H = I \cos \phi$.Rearranging for $\cos \phi$: $\cos \phi = \frac{B_H}{I}$.Substituting the values: $\cos \phi = \frac{0.3}{0.6} = \frac{1}{2}$.Since $\cos \phi = \frac{1}{2}$,the angle of dip is $\phi = 60^o$.

At a place,the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are $0.3 \; Oersted$ and $0.6 \; Oersted$ respectively. The value of the angle of dip at this place will be.....$^o$

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