Two bar magnets with magnetic moments $2M$ and $M$ are fastened together at right angles to each other at their centres to form a crossed system,which can rotate freely about a vertical axis through the centre. The crossed system sets in the Earth's magnetic field with the magnet having magnetic moment $2M$ making an angle $\theta$ with the magnetic meridian such that:

  • A
    $\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  • B
    $\theta = \tan^{-1}(\sqrt{3})$
  • C
    $\theta = \tan^{-1}\left(\frac{1}{2}\right)$
  • D
    $\theta = \tan^{-1}\left(\frac{3}{4}\right)$

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Similar Questions

At an angle of $30^{\circ}$ to the magnetic meridian,the apparent dip is $45^{\circ}$. Find the true dip.

The correct relation is:
$B_H$ = Horizontal component of earth's magnetic field; $B_V$ = Vertical component of earth's magnetic field and $B$ = Total intensity of earth's magnetic field.

Fill in the blanks:
$(i)$ The declination is ...... at higher latitudes.
$(ii)$ The declination in India is ...... .

The angle of dip is the angle

The line on the earth's surface joining the points where the magnetic field is horizontal is known as:

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