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 60o then the value of vertical component of earth's magnetic field will be-(in Wb/m2)-
Answer & explanation
Correct answer: option 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$