If the dip circle is set at $45^o$ to the magnetic meridian, then the apparent dip is $30^o$. The true dip. of the place is
${\tan ^{ - 1}}\,\left( {\frac{1}{{\sqrt 3 }}} \right)$
${\tan ^{ - 1}}\,\left( {\frac{1}{{\sqrt 6 }}} \right)$
${\tan ^{ - 1}}\,\left( {\frac{2}{{\sqrt 3 }}} \right)$
${\tan ^{ - 1}}\,\left( {\frac{{\sqrt 3 }}{2}} \right)$
A compass needle whose magnetic moment is $60 \,amp × m^2$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $40\, \mu Wb/m^2$, experiences a torque $1.2 \times {10^{ - 3}}\,N \times m.$ What is the declination at this place......$^o$
Assertion : If a compass needle be kept at magnetic north pole of the earth the compass needle may stay in any direction.
Reason : Dip needle will stay vertical at the north pole of earth
The relations amongst the three elements elements earth's magnettic fleld, namely horizontal component $H$, vertical component $V$ and dip $\delta$ are, $( B_{E}=$ total magnetic fleld)
Let $V $ and $H $ be the vertical and horizontal components of earth's magnetic field at any point on earth. Near the north pole
The vertical component of the earth's magnetic field is $6 \times 10^{-5} T$ at any place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be $\left(\right.$ Given $\left.\tan 37^{\circ}=\frac{3}{4}\right)$