At a place the earth's horizontal component of magnetic field is 0.36×10−4Weber/m2. If the angle of dip at that place is 60∘, then the vertical component of earth's field at that place in Weber/m2 will be approxmately

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At a place the earth's horizontal component of magnetic field

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 $60^{\circ}$ , then the vertical component of earth's field at that place in $Weber / m^{2}$ will be approxmately

0.12 \times 10^{- 4}

0.24 \times 10^{- 4}

0.40 \times 10^{- 4}

0.62 \times 10^{- 4}

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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^{\circ}

, then the vertical component of earth's field at that place in

Weber / m^{2}

will be approxmately

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Question Text	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 $60^{\circ}$ , then the vertical component of earth's field at that place in $Weber / m^{2}$ will be approxmately
Answer Type	Video solution: 1
Upvotes	151