Question

The vertical and horizontal component of earth's magnetic field at a place are $2\times {10}^{-5}T$ and $3.464\times {10}^{-5}T$ respectively. Calculate the angle of dip and resultant earth's magnetic field at that place.

• A. $\delta ={60}^{\circ }B=4\times {10}^{-5}T$.
• B. $\delta ={30}^{\circ }B=4\times {10}^{-5}T$.
• C. $\delta ={15}^{\circ }B=4\times {10}^{-5}T$.
• D. $\delta ={30}^{\circ }B=2\times {10}^{-5}T$.

Answer 1

Answer: (B)

$\delta ={30}^{\circ }B=4\times {10}^{-5}T$.

$B_V=2\times {10}^{-5}T{B}_H=3.464\times {10}^{-5}T\tan \delta =\frac{B_V}{B_H}\tan \delta =\frac{2\times {10}^{-5}T}{3.464\times {10}^{-5}T}\tan \delta =\frac{1}{1.732}\delta ={30}^{\circ }$

$B_H=B\cos \delta B=\frac{B_H}{\cos \delta }B=\frac{3.464\times {10}^{-5}}{\cos {30}^{\circ }}B=\frac{3.464\times {10}^{-5}}{\sqrt{3}/2}B=4\times {10}^{-5}T$

Hence option 'B' is correct.