Question:
At a given place the ratio of the total magnetic field of Earth to the horizontal component of Earth's magnetic field is \(2\). The angle of dip at that place is
At a given place the ratio of the total magnetic field of Earth to the horizontal component of Earth's magnetic field is \(2\). The angle of dip at that place is
Show Hint
Useful relations in Earth's magnetism:
\[
B_H=B\cos\delta,
\]
\[
B_V=B\sin\delta,
\]
\[
\tan\delta=\frac{B_V}{B_H}.
\]
If \(\dfrac{B}{B_H}=2\), then
\[
\sec\delta=2
\]
which directly gives
\[
\delta=60^\circ.
\]
Updated On: Jul 9, 2026
- \(30^\circ\)
- \(45^\circ\)
- \(60^\circ\)
- \(90^\circ\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
The horizontal component of Earth's magnetic field is related to the total magnetic field by
\[
B_H = B\cos\delta,
\]
where
\[
B=\text{total magnetic field},
\]
and
\[
\delta=\text{angle of dip}.
\]
Step 1: Use the given ratio. Given, \[ \frac{B}{B_H}=2. \] Substituting \[ B_H=B\cos\delta, \] we get \[ \frac{B}{B\cos\delta}=2. \] \[ \frac{1}{\cos\delta}=2. \] \[ \cos\delta=\frac{1}{2}. \]
Step 2: Find the angle of dip. \[ \delta=\cos^{-1}\left(\frac12\right). \] \[ \delta=60^\circ. \]
Step 3: Write the final answer. \[ \boxed{\delta=60^\circ} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Use the given ratio. Given, \[ \frac{B}{B_H}=2. \] Substituting \[ B_H=B\cos\delta, \] we get \[ \frac{B}{B\cos\delta}=2. \] \[ \frac{1}{\cos\delta}=2. \] \[ \cos\delta=\frac{1}{2}. \]
Step 2: Find the angle of dip. \[ \delta=\cos^{-1}\left(\frac12\right). \] \[ \delta=60^\circ. \]
Step 3: Write the final answer. \[ \boxed{\delta=60^\circ} \] \[ \boxed{\text{Answer = (C)}} \]
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