1
AIIMS 2018
MCQ (Single Correct Answer)
+1
-0.33

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

A
$$0.12 \times 10^{-4}$$
B
$$0.24 \times 10^{-4}$$
C
$$0.40 \times 10^{-4}$$
D
$$0.622 \times 10^{-4}$$
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