First slide

Magnetic elements of Earth

Question

If a magnet is suspended at an angle of $30^{0}$ to the magnetic meridian, the dip needle makes an angle of $60^{0}$ with the horizontal. The real dip is

Moderate

A

$\tan^{- 1} (3 / 2)$
B

$\tan^{- 1} (2 / \sqrt{3})$
C

$\tan^{- 1} (\sqrt{3} / 2)$
D

$\tan^{- 1} (2 / 3)$

Solution

$\tan θ = (B_{V} / B_{H})$ [real dip is $θ$ ]…………………(1)
In the place situated at $30^{0}$ with the magnetic meridian, the horizontal component of the
earth’s magnetic field will be $B_{H} \cos 30^{0}$ while the vertical component will be $B_{v}$ . The angle
of dip in this place is
$\tan 60^{o} = \frac{B_{v}}{B_{H} \cos 30^{o}}$
$\Rightarrow t an 60^{o} . \cos 30^{o} = \frac{B_{v}}{B_{H}}$
From equation (1),
$\Rightarrow \tan θ = \sqrt{3} . \frac{\sqrt{3}}{2}$
$\Rightarrow \tan θ = \frac{3}{2}$
$\Rightarrow θ = \tan^{- 1} (\frac{3}{2})$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App