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Question: If a magnetic is suspended at an angle \({30^ \circ }\) to the magnetic meridian, the dip needle mak...

If a magnetic is suspended at an angle 30{30^ \circ } to the magnetic meridian, the dip needle makes an angle of 45{45^ \circ } with the horizontal. The real dip is:
(A) tan1(32){\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)
(B) tan1(3){\tan ^{ - 1}}\left( {\sqrt 3 } \right)
(C) tan1(32){\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt 2 }}} \right)
(D) tan1(23){\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)

Explanation

Solution

Hint We should know that the magnetic meridian is defined as the equivalent imaginary line that connects the magnetic south and the north poles. The direction is taken to be a horizontal component of the magnetic force lines, which are along the surface of the earth. So, the compass needle will always be parallel to the magnetic meridian.

Complete step by step answer:
As the angle and its values are given at first we have to mention the values as per the question in the given answer.
We know that,
Angle of dip, δ=45\delta = {45^ \circ }
This gives us an idea about the angle that is made by the dip needle, and the value is mentioned as per the question.
So, we can write that:
δ=tanδcosθ  \delta = \dfrac{{\tan \delta }}{{\cos \theta }}\;
Now after putting the values we get that:
=tan45cos30=132=23= \dfrac{{\tan 45}}{{\cos {{30}^ \circ }}} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{2}{{\sqrt 3 }}
So, now we can say that:
Hence we can say that the value of the dip we get it as:
δ=tan1(23)\therefore \delta = {\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)

Hence, the correct answer is Option D.

Note In this question we have come across the term angle of dip. This is defined as the angle that is made by the earth’s magnetic field lines with the horizontal. Whenever the horizontal component and the vertical component of the magnetic field of the earth are equal to each other, the value of the angle of dip is 45 degrees.