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Question

(D) In the magnetic meridian,the apparent dip $\delta$ is related to the vertical component $B_V$ and horizontal component $B_H$ of the Earth's magnet

Vedclass · Accepted Answer

$1 / \cos \alpha$

Vedclass · Answer

(D) In the magnetic meridian,the apparent dip $\delta$ is related to the vertical component $B_V$ and horizontal component $B_H$ of the Earth's magnetic field by $	an \delta = \frac{B_V}{B_H}$.When the dip circle is rotated by an angle $\alpha$ in the horizontal plane,the effective horizontal component becomes $B_H^{\prime} = B_H \cos \alpha$,while the vertical component $B_V$ remains unchanged.The new apparent angle of dip $\delta^{\prime}$ is given by $	an \delta^{\prime} = \frac{B_V}{B_H^{\prime}} = \frac{B_V}{B_H \cos \alpha}$.Substituting $	an \delta = \frac{B_V}{B_H}$,we get $	an \delta^{\prime} = \frac{	an \delta}{\cos \alpha}$.Therefore,the ratio $\frac{	an \delta^{\prime}}{	an \delta} = \frac{1}{\cos \alpha}$.

$A$ dip circle lies initially in the magnetic meridian,it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

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