Question

The horizontal component of the earth's magnetic field at any place is $0.36\times 10^{- 4} \, Wb \, m^{- 2}$ . If the angle of dip at that place is $60^\circ $ then the value of the vertical component of earth's magnetic field will be (in $Wb \, m^{- 2}$ )

• A. $0.12\times 10^{- 4}$
• B. $0.24\times 10^{- 4}$
• C. $0.40\times 10^{- 4}$
• D. $0.62\times 10^{- 4}$

Answer 1

Answer: (D)

Given,
$B_{H}=0.36\times 10^{- 4}Wbm^{- 2}$
Angle of dip
$tan\theta =\frac{B_{V}}{B_{H}}$
$tan60^\circ =\frac{B_{V}}{0.36 \times 1 0^{- 4}}$
$B_{V}=\left(0.36\right)\left(\sqrt{3}\right)\times 10^{- 4}$
$B_{V}=0.62\times 10^{- 4}Wbm^{- 2}$