# If a magnet is suspended at an angle of 30 to the earth magnetic meridian , the dip needle makes angle of 45 with the horizontal . the real dip is ?

If a magnet is suspended at an angle of ${30}^{\circ }$ to the earth magnetic meridian , the dip needle makes angle of ${45}^{\circ }$ with the horizontal . the real dip is ?
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pawia6g
Let the vertical component of earth's magnetic field be V and its horizontal component at magnetic meridian be H.
Then the angle of dip $\theta$ at magnetic meridian will be given by
$\mathrm{tan}\theta =\frac{V}{H}$...[1]
When the dip needle is suspended at an angle of ${30}^{\circ }$ to the earth magnetic meridian then it makes an angle ${45}^{\circ }$ with the horizontal.
In this situation vertical component of earth's field responsible for its orientation with the horizontal direction, remains same as V but the component in horizontal direction becomes $H\mathrm{cos}{30}^{\circ }$
So $\mathrm{tan}{45}^{\circ }=\frac{V}{H\mathrm{cos}{30}^{\circ }}$
$⇒\frac{V}{H}=\mathrm{cos}{30}^{\circ }$...[2]
Comparing [1] and [2] we get
$\mathrm{tan}\theta =\mathrm{cos}{30}^{\circ }=\frac{\sqrt{3}}{2}$
$⇒\theta ={\mathrm{tan}}^{-1}\left(\frac{\sqrt{3}}{2}\right)\approx {40.9}^{\circ }$