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

Question

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

1 Answer

Impact of this question

44015 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-16T04:03:02

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

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 $θ$ $theta$ at magnetic meridian will be given by

$tantheta=V/H......[1]$

When the dip needle is suspended at an angle of $30^@$ to the earth magnetic meridian then it makes an angle $45^@$ 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 $Hcos30^@$

So $tan45^@=V/(Hcos30^@)$

$=>V/H=cos30^@.......[2]$

Comparing [1] and [2] we get

$tantheta=cos30^@=sqrt3/2$

$=>theta=tan^-1(sqrt3/2)~~40.9^@$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If a magnet is suspended at an angle of 30^o30o30^o to the earth magnetic meridian , the dip needle makes angle of 45^o45o45^o with the horizontal . the real dip is ?

1 Answer

Explanation:

Related questions

Impact of this question

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