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…..

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

Author: Rajesh Jha

QuantumStudy.com

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