The horizontal component of earth's magnetic field at any place is 0.36×10^–4 Weber/m². If the angle of dip at that place is 60^o then the value of vertical component of earth's magnetic field will be-(in Wb/m²)-

$0.62 \times 10^{-4}$

$ B_V = B_H tan \theta = 0.36\times 10^{-4} tan 60^o = 0.36\times 10^{-4} \sqrt 3 = 0.62 \times 10^{-4}T$