Question

Consider the earth as a short magnet with the centre coinciding with the centre of the earth and dipole moment $M$. The angle of dip $\theta$ at the place where the magnetic latitude is $\lambda$ is :

• A. ${\tan }^{-1}\left(2\tan \lambda \right)$
• B. ${\tan }^{-1}\left(\tan \lambda \right)$
• C. ${\tan }^{-1}\left(\tan \lambda /2\right)$
• D. ${\tan }^{-1}\left(4\tan \lambda \right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(2\tan \lambda \right)$

In below solution angle of dip is taken as $\delta$,

$B_r=\frac{2{\mu }_{o}Mcos\theta }{4\pi r^3}$

$B_{\theta }=\frac{{\mu }_{o}Msin\theta }{4\pi r^3}$ as $\theta =90+\lambda$

$B_r=\frac{2{\mu }_{o}Mcos(90+\lambda )}{4\pi r^3}=\frac{-2{\mu }_{o}Msin\lambda }{4\pi r^3}$

$B_{\theta }=\frac{{\mu }_{o}Msin(90+\lambda )}{4\pi r^3}=\frac{{\mu }_{o}Mcos\lambda }{4\pi r^3}$

dividing the two equations we get,

$\frac{B_r}{B_{\theta }}=-2tan\lambda$ or $tan\delta =2tan\lambda$

$\therefore$ Angle of dip is $=\delta =ta{n}^{-1}\left(2tan\lambda \right)$