Question

If $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^\circ$, then the total magnetic field at that place is

• A. $\sqrt{5} B_H$
• B. $\sqrt{3} B_V$
• C. $\frac{2}{\sqrt{3}} B_V$
• D. $\frac{\sqrt{3}}{2} B_H$

Hint:

Remember the triangle of magnetic components: $B$ is the hypotenuse, $B_H$ is adjacent to $\delta$, $B_V$ is opposite to $\delta$.

Answer 1

The Correct Option is C

Step 1: Formulae:
Angle of dip $\delta = 60^\circ$. Vertical component $B_V = B \sin \delta$. Horizontal component $B_H = B \cos \delta$. Total magnetic field $B$.
Step 2: Express B in terms of components:
From the vertical component equation: $B = \frac{B_V}{\sin \delta}$.
Step 3: Substitute $\delta$:
$B = \frac{B_V}{\sin 60^\circ} = \frac{B_V}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} B_V$. (Checking horizontal option: $B = \frac{B_H}{\cos 60^\circ} = 2B_H$. This is not among the options in that form).