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(A) In the magnetic meridian,the vertical component of the Earth's magnetic field is $B_V$ and the horizontal component is $B_H$. The angle of dip $

Vedclass · Accepted Answer

$\frac{1}{\cos x}$

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(A) In the magnetic meridian,the vertical component of the Earth's magnetic field is $B_V$ and the horizontal component is $B_H$. The angle of dip $	heta$ is given by:$	an 	heta = \frac{B_V}{B_H}$ ..... $(i)$When the dip circle is rotated by an angle $x$ in the horizontal plane,the effective horizontal component becomes $B_H' = B_H \cos x$,while the vertical component $B_V$ remains unchanged.The new angle of dip $	heta'$ is given by:$	an 	heta' = \frac{B_V}{B_H \cos x}$ ..... $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{	an 	heta'}{	an 	heta} = \frac{B_V / (B_H \cos x)}{B_V / B_H} = \frac{B_V}{B_H \cos x} 	imes \frac{B_H}{B_V} = \frac{1}{\cos x}$

$A$ dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle $x$ in the horizontal plane and then it shows an angle of dip $\theta'$. Then $\frac{\tan \theta'}{\tan \theta}$ is

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