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

##### 1 Answer
Apr 16, 2017

Real dip $\theta = {40.9}^{\circ}$

#### Explanation:

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

$\tan \theta = \frac{V}{H} \ldots \ldots \left[1\right]$

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 $H \cos {30}^{\circ}$

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

$\implies \frac{V}{H} = \cos {30}^{\circ} \ldots \ldots . \left[2\right]$

Comparing [1] and [2] we get

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

=>theta=tan^-1(sqrt3/2)~~40.9^@
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}