magnetism

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 $60^o$, then the vertical component of earth's field at that place in weber/m2 will be approximately

1) $0.12 \times {10^{ - 4}}$

2) $0.24 \times {10^{ - 4}}$

3) $0.40 \times {10^{ - 4}}$

4) $0.62 \times {10^{ - 4}}$

Solution:

Explanation: No Explanation

Earth Magnetism

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