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:

$8×10^{-4} T$

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,

$B_H=4×10^{-4}T$

$\phi=60°$

$∴B=\frac{B_H}{\cos 60}=\frac{4×10^{-4}}{1/2}$

$=8×10^{-4}T$