Question

At a place the earth's horizontal component of magnetic field is $0.36\times {10}^{-4}Weber/m^2$ . If the angle of dip at that place is ${60}^0$, then the vertical component of earth's field at that place in $Weber/m^2$ will be approximately

• A. $0.62\times \times 0^{-4}$
• B. $0.24\times \times 0^{-4}$
• C. $0.12\times \times 0^{-4}$
• D. $0.40\times \times 0^{-4}$

Answer 1

The correct option is A $0.62\times \times 0^{-4}$

Let $B_H$ and $B_V$ be the horizontal and vertical components of the magnetic field.

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$\therefore \frac{B_v}{B_H}=\frac{Bsin\varphi }{Bcos\varphi }=tan\varphi$

Where, $\varphi$ is the angle of the dip.

$\therefore B_V=B_{H}tan\varphi$

$=0.36\times {10}^{-4}Weber/m^2\times tan{60}^0$
$=0.623\times {10}^{-4}wb/m^2$

Hence, correct option is $\left(d\right)$.