At a certain place, the angle of dip is $60^\circ$ and the horizontal component of the Earth's magnetic field is $0.32\times10^{-4}\ \text{T}$. Calculate the vertical component and the total magnitude of the Earth's magnetic field at that place.

At a certain place, the angle of dip is 60^° and the horizontal component of the Earth's magnetic field is 0.3

Numerical · Q25

60^\circ

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Given $\delta=60^\circ$ , $H=0.32\times10^{-4}\ \text{T}$ ; $\tan60^\circ\approx1.732$ , $\cos60^\circ=0.5$ .

Vertical component:

V = H\tan\delta = (0.32\times10^{-4})\times1.732 \approx 5.54\times10^{-5}\ \text{T}

Total field:

$B_E = \frac{H}{\cos\delta} = \frac{0.32\times10^{-4}}{0.5} = 0.64\times10^{-4} = 6.4\times10^{-5}\ \text{T}$ …