
The true value of dip at a place is ${45^\circ }$. If the plane of the dip circle is turned through ${60^\circ }$ from the magnetic meridian, what will be the apparent angle of dip?
Answer
217.8k+ views
Hint: We will learn about the magnetic angle of dip, angle of the magnetic meridian, and apparent angle of magnetic dip. We can then figure out their relations and determine the apparent angle of dip at the point in our problem.
Formula used:
$\tan \delta ' = \dfrac{{\tan \delta }}{{\cos \theta }}$
Complete step by step solution:
The angle made by the Earth's magnetic field with the horizontal at any point measures the magnetic angle of dip at the point. The dip angle is produced because the magnetic lines of the Earth are not parallel to its surface. That is why a magnetic needle's north pole will point downward in the northern hemisphere and point upward in the southern hemisphere. Moreover, the south pole of the magnetic needle will behave exactly the opposite.
The magnetic dip at any place is determined by a dip circle. If this imaginary circle's plane is not in the magnetic meridian, it will not show Earth's magnetic field's exact direction.
And we have,
$\Rightarrow \tan \delta ' = \dfrac{{\tan \delta }}{{\cos \theta }}$
where,
$\delta $ is the angle of dip
$\delta '$ is the apparent angle of dip
$\theta $ is the angle made by the dip circle
Now we put $\delta = {45^ \circ }$ and $\theta = {60^ \circ }$ in the above equation, and we get-
$\Rightarrow \tan \delta ' = \dfrac{{\tan \left( {{{45}^ \circ }} \right)}}{{\cos \left( {{{60}^ \circ }} \right)}}$
$ \Rightarrow \tan \delta ' = \dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}$
Upon solving further we get,
$\Rightarrow \tan \delta ' = \dfrac{2}{{\sqrt 2 }}$
$ \Rightarrow \tan \delta ' = \sqrt 2 $
From this, we get the apparent angle of dip-
$\Rightarrow \delta ' = {\tan ^{ - 1}}\left( {\sqrt 2 } \right)$
$ \Rightarrow \delta ' = {54.74^ \circ }$
Note: The angle measured downwards from the horizontal line is called a positive dip, and the angle measured upwards from the horizontal line is called a negative dip. The dip angle measured in a vertical plane that is not perpendicular to the strike line is termed as the apparent angle of dip. We can also measure the true dip angle if the strike is known.
Formula used:
$\tan \delta ' = \dfrac{{\tan \delta }}{{\cos \theta }}$
Complete step by step solution:
The angle made by the Earth's magnetic field with the horizontal at any point measures the magnetic angle of dip at the point. The dip angle is produced because the magnetic lines of the Earth are not parallel to its surface. That is why a magnetic needle's north pole will point downward in the northern hemisphere and point upward in the southern hemisphere. Moreover, the south pole of the magnetic needle will behave exactly the opposite.
The magnetic dip at any place is determined by a dip circle. If this imaginary circle's plane is not in the magnetic meridian, it will not show Earth's magnetic field's exact direction.
And we have,
$\Rightarrow \tan \delta ' = \dfrac{{\tan \delta }}{{\cos \theta }}$
where,
$\delta $ is the angle of dip
$\delta '$ is the apparent angle of dip
$\theta $ is the angle made by the dip circle
Now we put $\delta = {45^ \circ }$ and $\theta = {60^ \circ }$ in the above equation, and we get-
$\Rightarrow \tan \delta ' = \dfrac{{\tan \left( {{{45}^ \circ }} \right)}}{{\cos \left( {{{60}^ \circ }} \right)}}$
$ \Rightarrow \tan \delta ' = \dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}$
Upon solving further we get,
$\Rightarrow \tan \delta ' = \dfrac{2}{{\sqrt 2 }}$
$ \Rightarrow \tan \delta ' = \sqrt 2 $
From this, we get the apparent angle of dip-
$\Rightarrow \delta ' = {\tan ^{ - 1}}\left( {\sqrt 2 } \right)$
$ \Rightarrow \delta ' = {54.74^ \circ }$
Note: The angle measured downwards from the horizontal line is called a positive dip, and the angle measured upwards from the horizontal line is called a negative dip. The dip angle measured in a vertical plane that is not perpendicular to the strike line is termed as the apparent angle of dip. We can also measure the true dip angle if the strike is known.
Recently Updated Pages
Arithmetic, Geometric & Harmonic Progressions Explained

Cartesian Form of Vector Explained: Formula, Examples & Uses

Apparent Frequency Explained: Formula, Uses & Examples

Calorimetry: Definition, Principles & Calculations

Centrifugal Force Explained: Definition, Formula & Examples

Charge in a Magnetic Field: Definition, Formula & Examples

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding Average and RMS Value in Electrical Circuits

