Exam

JEE Advanced

Subject

Physics

Chapter

Magnetic Properties of Matter

JEE Advanced · Physics · Magnetic Properties of Matter

The earth is considered as a short magnet with its centre...

The earth is considered as a short magnet with its centre coinciding with the geometric centre of earth. The angle of $\operatorname{dip} \phi$ related to the magnetic latitude $\alpha \lambda$ as

Step-by-step solution

For a dipole at position $(\mathrm{R}, \mathrm{Q})$
$\mathrm{B}_{\mathrm{R}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \mathrm{M} \cos \theta}{\mathrm{R}^{3}}$ ...(i)
and $\mathrm{B}_{\mathrm{Q}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{\mathrm{M} \sin \theta}{\mathrm{R}^{3}}$ ...(ii)
Also $\tan \phi=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=-\frac{\mathrm{B}_{\mathrm{R}}}{\mathrm{B}_{\mathrm{Q}}}$ ...(iii)


Dividing eq. (i) by (ii)
$\frac{\mathrm{B}_{\mathrm{R}}}{\mathrm{B}_{\mathrm{Q}}}=\frac{2 \cos \theta}{\sin \theta}=2 \cot \theta$
From eq. (iii) and (iv)
$\tan \phi=-2 \cot \theta$
From figure, $\theta=90^{{}^{{}^{\circ}}}+\lambda$
$\therefore \quad \tan \phi=-2 \cot (90+\lambda)$
$\tan \phi=2 \tan \lambda$