Horizontal component of earth’s magnetic field at a place is \(\sqrt

1.	Horizontal component of earth’s magnetic field at a place is \(\sqrt 3\) times the vertical component. What is the value of inclination at that place?
Answer» Given that: B_H = \(\sqrt3\) B_v Also, B_H = B cos\(\delta\) B_v= B sin\(\delta\) ∴ tan\(\delta\) = \(\frac{B_v}{B_H}\) = \(\frac{B_v}{\sqrt{3}\,B_v}\) = \(\frac{1}{\sqrt3}\) \(\Rightarrow\) \(\delta\) = 30º \(\frac{Bv}{B_H}=tan\theta\) Putting values,\(\theta\)=30°

Horizontal component of earth’s magnetic field at a place is \(\sqrt 3\) times the vertical component. What is the value of inclination at that place?

Answer»

Given that:

B_H = \(\sqrt3\) B_v

Also, B_H = B cos\(\delta\)

B_v= B sin\(\delta\)

∴ tan\(\delta\) = \(\frac{B_v}{B_H}\) = \(\frac{B_v}{\sqrt{3}\,B_v}\) = \(\frac{1}{\sqrt3}\)

\(\Rightarrow\) \(\delta\) = 30º

\(\frac{Bv}{B_H}=tan\theta\)

Putting values,\(\theta\)=30°

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Horizontal component of earth’s magnetic field at a place is \(\sqrt 3\) times the vertical component. What is the value of inclination at that place?

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