Question

At a certain place in the magnetic meridian, the earth's magnetic field is twice its vertical component. The ratio of horizontal component of earth's magnetic field and the total magnetic field of the earth at that place is

• A. $\sqrt{3}:2$
• B. 1:2
• C. 1:$\sqrt{3}$
• D. 1:3

Hint:

The components of the Earth's magnetic field are related by $B^2 = B_H^2 + B_V^2$. For problems involving ratios, substitute one component into the main equation to find the ratio of the two components, and then calculate the final required ratio.

Answer 1

The Correct Option is A

Step 1: Relate the components of the Earth's magnetic field.
Let $B$ be the total intensity of the Earth's magnetic field. Let $B_H$ be the horizontal component and $B_V$ be the vertical component. The total field is the vector sum of its components: \[ B^2 = B_H^2 + B_V^2. \] The angle of dip, $\delta$, is the angle between the total field and the horizontal component, defined by: \[ \tan\delta = \frac{B_V}{B_H}. \]

Step 2: Use the given condition to find the ratio of components.
We are given that the Earth's magnetic field is twice its vertical component: \[ B = 2 B_V. \] Substitute this into the total field equation: \[ (2 B_V)^2 = B_H^2 + B_V^2. \] \[ 4 B_V^2 = B_H^2 + B_V^2. \] \[ B_H^2 = 4 B_V^2 - B_V^2 = 3 B_V^2. \]

Step 3: Find the relationship between $B_H$ and $B_V$.
Take the square root of the equation from Step 2: \[ B_H = \sqrt{3} B_V. \]

Step 4: Calculate the required ratio.
The question asks for the ratio of the horizontal component to the total magnetic field, $\frac{B_H}{B}$. From the given information, we have $B = 2 B_V$ and $B_V = \frac{1}{\sqrt{3}} B_H$. Substitute $B_V$ in the first relation: \[ B = 2 \left(\frac{1}{\sqrt{3}} B_H\right) = \frac{2}{\sqrt{3}} B_H. \] Rearrange to find the ratio $\frac{B_H}{B}$: \[ \frac{B_H}{B} = \frac{\sqrt{3}}{2}. \] The ratio is $\sqrt{3}:2$.