The Earth's magnetic field at a certain place has a horizontal component of $0.3 \ G$ and the total strength of $0.5 \ G$. The angle of dip is:

  • A
    $\tan^{-1} \left( \frac{3}{4} \right)$
  • B
    $\sin^{-1} \left( \frac{3}{4} \right)$
  • C
    $\tan^{-1} \left( \frac{4}{3} \right)$
  • D
    $\sin^{-1} \left( \frac{3}{5} \right)$

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Similar Questions

The vertical component of the Earth's magnetic field is zero at:

$A$ dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position,the angle of dip is $40^{\circ}$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $30^{\circ}$ with the magnetic meridian. In this position,the needle will dip by an angle:

The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$

At a certain place,the vertical component of the Earth's magnetic field is $\sqrt{3}$ times the horizontal component of the Earth's magnetic field. If a magnetic needle is suspended freely in the air,then it will incline:

If the angle of dip at places $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively,the ratio of the horizontal component of the Earth's magnetic field at $A$ to that at $B$ will be.
$[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 30^{\circ}=\frac{1}{2}, \quad \cos 30^{\circ}=\frac{\sqrt{3}}{2}]$

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