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 ?

c0nman56
2022-10-21
Answered

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

Answered 2022-10-22
Author has **14** answers

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 $\mathrm{tan}{45}^{\circ}=\frac{V}{H\mathrm{cos}{30}^{\circ}}$

$\Rightarrow \frac{V}{H}=\mathrm{cos}{30}^{\circ}$...[2]

Comparing [1] and [2] we get

$\mathrm{tan}\theta =\mathrm{cos}{30}^{\circ}=\frac{\sqrt{3}}{2}$

$\Rightarrow \theta ={\mathrm{tan}}^{-1}(\frac{\sqrt{3}}{2})\approx {40.9}^{\circ}$

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 $\mathrm{tan}{45}^{\circ}=\frac{V}{H\mathrm{cos}{30}^{\circ}}$

$\Rightarrow \frac{V}{H}=\mathrm{cos}{30}^{\circ}$...[2]

Comparing [1] and [2] we get

$\mathrm{tan}\theta =\mathrm{cos}{30}^{\circ}=\frac{\sqrt{3}}{2}$

$\Rightarrow \theta ={\mathrm{tan}}^{-1}(\frac{\sqrt{3}}{2})\approx {40.9}^{\circ}$

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