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Questions  >>  CBSE XII  >>  Physics  >>  Magnetism and matter
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Q)

The vertical component of earth magnetic field at a plane is $\large\frac{1}{\sqrt 3}$ times the horizontal component. What is the value of angle of dip at this place ?

$\begin{array}{1 1} 30^{\circ} \\ 60^{\circ} \\ 90^{\circ} \\ 0^{\circ} \end{array} $

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A)
Solution :
$B_v= \large\frac{1}{\sqrt 3}$$B_H$
$\tan \delta = \large\frac{B_v}{B_H}$
$\tan \delta = \large\frac{1}{\sqrt 3}$
$\delta = 30^{\circ}$
Answer : $30^{\circ}$
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