First slide

Magnetic elements of Earth

Question

A compass needle placed at a distance ‘r’ from a short magnet in tan A position shows a deflection of $60^{0}$ , if the distance is increased to $r {(3)}^{1 / 3}$ . Then the deflection of the compass needle is

Moderate

A

$30^{0}$
B

$0^{0}$
C

$45^{0}$
D

$90^{0}$

Solution

The magnetic field at a point along the axial line of a bar magnet is perpendicular to the horizontal component of Earth's magnetic field. If a magnetometer and a bar magnet are placed in such way that this condition is satisfied, then this arrangement is called Tan A position.
In the $tan A$ position
$\tan θ = \frac{B_{a x i a l}}{B_{H}}$
$\Rightarrow \frac{μ_{0}}{4 π} . \frac{2 M}{r^{3}} = B_{H} \tan θ$
$\Rightarrow {(\frac{r_{1}}{r_{2}})}^{3} = \frac{\tan θ_{2}}{\tan θ_{1}}$
$G i v e n : (\frac{r_{1}}{r_{2}}) = \frac{r}{r {(3)}^{1 / 3}} = \frac{1}{3^{1 / 3}}$
$∴ \tan θ_{2} = (\tan θ_{1}) {(\frac{r_{1}}{r_{2}})}^{3}$
$\Rightarrow \tan θ_{2} = (\tan 60^{0}) {(\frac{1}{3^{1 / 3}})}^{3}$
$\Rightarrow \tan θ_{2} = \sqrt{3} . \frac{1}{3} = \frac{1}{\sqrt{3}}$
$\Rightarrow θ_{2} = 30^{0}$ .

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