The vertical component of earth-s magnetic field a place is sqrt-3- times the horizontal component the value of angle of dip this place is-45-o60-o90-o30-o

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Toppr Admin · Accepted Answer

As- BV-x221A-3BHAlso- tan-x3B4-BVBH-x221A-3BHBH-x221A-3-xA0-or -x3B4-tan-x2212-1-x221A-3-60o-x2234- Angle of dip- -x3B4-60o-

The vertical component of earth's magnetic field at a place is $\sqrt{3}$ times the horizontal component the value of angle of dip at this place is?
$45^{o}$
$60^{o}$
$90^{o}$
$30^{o}$

As, $B_{V} = \sqrt{3} B_{H}$

Also, $tan δ = \frac{B_{V}}{B_{H}} = \frac{\sqrt{3} B_{H}}{B_{H}} = \sqrt{3}$

or $δ = {tan}^{- 1} (\sqrt{3}) = 60^{o}$

$∴$ Angle of dip, $δ = 60^{o}$ .

The vertical component of earth's magnetic field at a place is √3 times the horizontal component the value of angle of dip at this place is?45o60o90o30o

As, BV=√3BHAlso, tanδ=BVBH=√3BHBH=√3 or δ=tan−1(√3)=60o∴ Angle of dip, δ=60o.

The vertical component of earth's magnetic field at a place is $\sqrt{3}$ times the horizontal component the value of angle of dip at this place is?
$45^{o}$
$60^{o}$
$90^{o}$
$30^{o}$

As, $B_{V} = \sqrt{3} B_{H}$

Also, $tan δ = \frac{B_{V}}{B_{H}} = \frac{\sqrt{3} B_{H}}{B_{H}} = \sqrt{3}$

or $δ = {tan}^{- 1} (\sqrt{3}) = 60^{o}$

$∴$ Angle of dip, $δ = 60^{o}$ .