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: BH = 0.36 × 10-4 weber/m² ; BV = ? ; 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} $
BV = 3.6√3 × 10-5 T