Question

If the horizontal component of earth's magnetic field at a place is $2.8 \times 10^{-5}$ T and the magnetic field of the earth at this place is $5.6 \times 10^{-5}$ T, then the inclination at the place is:

• A. $45^\circ$
• B. $30^\circ$
• C. $60^\circ$
• D. $90^\circ$

Hint:

Inclination (or dip) is the angle the total magnetic field makes with the horizontal component.

Answer 1

The Correct Option is C

Step 1: Note the given fields.
The horizontal part of earth's magnetic field is $B_H = 2.8\times10^{-5}\ T$. The total field of the earth at that place is $B = 5.6\times10^{-5}\ T$. We must find the angle of dip (inclination).

Step 2: Recall what dip means.
The total field of the earth points at an angle below the horizontal. This angle is the angle of dip $\delta$. The horizontal field is just the flat part of that total field.

Step 3: Write the relation.
The horizontal component is the total field times the cosine of the dip: $B_H = B\cos\delta$.

Step 4: Rearrange for the angle.
So $\cos\delta = \dfrac{B_H}{B}$.

Step 5: Put in the numbers.
$\cos\delta = \dfrac{2.8\times10^{-5}}{5.6\times10^{-5}} = 0.5$.

Step 6: Find the angle.
The angle whose cosine is $0.5$ is $60^\circ$. So the inclination at that place is $60^\circ$. \[ \boxed{60^\circ} \]