Q. The horizontal component of earth’s magnetic field at place is 0.36× 10^{-4} weber/m². If the angle of dip at that placed is 60° then the value of vertical component of earth’s magnetic field will be (in wb/m²)

(a) 6 × 10^{-5} T

(b) 6√2 × 10^{-5} T

(c) 3.6√3 × 10^{-5} T

(d) √2 × 10^{-5} T

Ans: (c)

Sol: B_{H} = 0.36 × 10^{-4} weber/m² ; B_{V} = ? ; δ = 60°

$ \displaystyle tan\delta = \frac{B_V}{B_H} $

$\displaystyle B_V = B_H tan\delta $

$ \displaystyle B_V = 0.36 \times 10^{-4} tan60 $

$\displaystyle B_V = 0.36 \times 10^{-4} \sqrt{3} $

B_{V} = 3.6√3 × 10^{-5} T