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GUJCET · Physics · Magnetism And Matter
At the place where dip angle is \(60^{\circ}\), what is the relation between vertical component \(Z_E\) and horizontal component \(H _{ E }\) of the earth magnetic field?
- A \(Z _{ E }= H _{ E }\)
- B \(Z _{ E }=\sqrt{3} H _{ E }\)
- C \(H _{ E }=\sqrt{3} Z _{ E }\)
- D \(H _{ E }=\frac{\sqrt{3}}{2} Z _{ E }\)
Answer & Solution
Correct Answer
(B) \(Z _{ E }=\sqrt{3} H _{ E }\)
Step-by-step Solution
Detailed explanation
B
\(\tan I =\frac{ Z _{ E }}{ H _{ E }}\)
\(\begin{array}{ll}\therefore \tan 60^{\circ}=\frac{ Z _{ E }}{ H _{ E }} \\ \therefore \sqrt{3}=\frac{ Z _{ E }}{ H _{ E }}\end{array}\)
\(Z _{ E }=\sqrt{3} H _{ E }\)
\(\tan I =\frac{ Z _{ E }}{ H _{ E }}\)
\(\begin{array}{ll}\therefore \tan 60^{\circ}=\frac{ Z _{ E }}{ H _{ E }} \\ \therefore \sqrt{3}=\frac{ Z _{ E }}{ H _{ E }}\end{array}\)
\(Z _{ E }=\sqrt{3} H _{ E }\)
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