Question

If the vertical component of the earth's magnetic field is 0.45 G at a location, and angle of dip is \( 60^\circ \), then magnetic field of earth at that location is:

• A. \( 0.26 \, {G} \)
• B. \( 0.52 \, {G} \)
• C. \( 0.3 \, {G} \)
• D. \( 0.7 \, {G} \)

Hint:

To calculate the total magnetic field when the vertical component and angle of dip are known, use the formula \( B_v = B \sin \delta \) and solve for \( B \).

Answer 1

The Correct Option is B

To find the magnetic field of the Earth at the location, we can use the following formula:

\( B_v = B \sin \delta \)

Where:

• \( B_v \) is the vertical component of the magnetic field, given as 0.45 G.
• \( B \) is the total magnetic field, which we need to find.
• \( \delta \) is the angle of dip, given as \( 60^\circ \).

Rearranging the formula to solve for \( B \), we have:

\( B = \frac{B_v}{\sin \delta} \)

Substituting the known values:

\( B = \frac{0.45}{\sin 60^\circ} \)

Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \),

\( B = \frac{0.45}{0.866} \approx 0.52 \, \text{G} \)

Thus, the magnetic field of the Earth at that location is approximately 0.52 G.

Answer 2

We are given that the vertical component \( B_v = 0.45 \, {G} \) and the angle of dip \( \delta = 60^\circ \). The total magnetic field \( B \) is related to the vertical component \( B_v \) and the angle of dip \( \delta \) by the formula: \[ B_v = B \sin \delta. \] Step 1:
Substitute the given values into the equation: \[ 0.45 = B \sin 60^\circ. \] Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we have: \[ 0.45 = B \times \frac{\sqrt{3}}{2}. \] Step 2:
Solve for \( B \): \[ B = \frac{0.45 \times 2}{\sqrt{3}} = \frac{0.9}{\sqrt{3}}. \] Step 3:
Now calculate the value of \( B \): \[ B = \frac{0.9}{1.732} \approx 0.52 \, {G}. \] Thus, the total magnetic field at that location is \( 0.52 \, {G} \).