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

In India,the magnetic declination at Delhi is . . . . . . .

The true value of the angle of dip at a place is $60^o$. The apparent dip in a plane inclined at an angle of $30^o$ with the magnetic meridian is:

The angle between the Earth's magnetic axis and the Earth's geographical axis is approximately $... ^\circ$.

If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then:

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Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_v = \text{vertical component of magnetic field} = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \text{horizontal component of magnetic field} = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
where $\theta = 90^\circ - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which the dip angle is zero.

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