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(A) Let $B_H$ be the horizontal component and $B_V$ be the vertical component of the Earth's magnetic field.Given that $B_H = B_V$.The angle of dip $\

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$\frac{\pi}{4}$

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(A) Let $B_H$ be the horizontal component and $B_V$ be the vertical component of the Earth's magnetic field.Given that $B_H = B_V$.The angle of dip $\delta$ is defined by the relation $	an(\delta) = \frac{B_V}{B_H}$.Substituting the given condition $B_H = B_V$ into the formula,we get $	an(\delta) = \frac{B_V}{B_V} = 1$.Since $	an(\delta) = 1$,the angle of dip $\delta = 45^\circ$.Converting degrees to radians,$45^\circ = \frac{\pi}{4}$ radians.

At a point on the surface of the Earth,the value of the horizontal component of the Earth's magnetic field is equal to the value of the vertical component of the Earth's magnetic field. The angle of dip is:

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