Question

The earth's magnetic field at a certain place has a horizontal component of $0.3G$ and total strength $0.5G$. Find angle of dip in ${\tan }^{-1}$.

• A. $\delta =ta{n}^{-1}\frac{4}{3}$
• B. $\delta =ta{n}^{-1}\frac{3}{4}$
• C. $\delta =ta{n}^{-1}\frac{5}{3}$
• D. $\delta =ta{n}^{-1}\frac{3}{5}$

Answer 1

Answer: (A)

$\delta =ta{n}^{-1}\frac{4}{3}$

Given that,
Total strength of field $B=0.5G$,
Horizontal component of the field $H=0.3G$,
Total strength of field is given by
$B=\sqrt{(H^2+V^2)}$
where, V is the vertical component of the field.
$B^2=H^2+V^2$
$V^2=B^2-H^2=(0.5{)}^2-(0.3{)}^2=0.4G$
Now, angle of dip is
$\tan \delta =\frac{V}{H}=\frac{0.4}{0.3}=\frac{4}{3}$
$\delta ={\tan }^{-1}\frac{4}{3}$