At a place, the horizontal component of earth's magnetic field is $4 × 10^{-4}T$ and angle of dip is 60°. The value of horizontal component of earth's magnetic field at the equator is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $8×10^{-4} T$
The horizontal component of Earth's magnetic field ($B_H$) is related to the total magnetic field (B) and Angle of Dip $(\phi)$ by -
$B_H=B\cos\phi$
where,
$B_H=4×10^{-4}T$
$\phi=60°$
$∴B=\frac{B_H}{\cos 60}=\frac{4×10^{-4}}{1/2}$
$=8×10^{-4}T$