Question

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

• A. $tan\varphi =\frac{1}{2tan\alpha }$
• B. $tan\lambda =tan\varphi$
• C. $tan\lambda =2tan\varphi$
• D. $tan\varphi =2tan\lambda$

Answer 1

The correct option is D $tan\varphi =2tan\lambda$

For a dipole at position (R, Q)
$B_R=\frac{{\mu }_0}{4\pi }.\frac{2Mcos\theta }{R^3}$ ....... (i)
and $B_Q=\frac{{\mu }_0}{4\pi }.\frac{Msin\theta }{R^3}$ ...... (ii)
Also $tan\varphi =\frac{B_V}{B_H}=-\frac{B_R}{B_Q}$ ......... (iii)
Dividing eeq. (i) by (ii)
$\frac{B_R}{B_Q}=\frac{2cos\theta }{sin\theta }=2cot\theta$ ........ (iv)
From eq. (iii) and (iv)
$tan\varphi =-2cot\theta$
From figure, $\theta ={90}^o+\lambda$
$\therefore tan\varphi =-2cot(90+\lambda )$
$tan\varphi =2tan\lambda$