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If a magnet is suspended at an angle of ...

If a magnet is suspended at an angle of `30^(@)` to the magnetic meridián, the dip needle makes an angle of `45^(@)` with the horizontal. The real dip is

A

`tan^(-1)(sqrt3"/"2)`

B

`tan^(-1)(sqrt3)`

C

`tan^(-1)(sqrt3"/"sqrt2)`

D

`tan^(-1)(2"/"sqrt3)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let the actual angle of dip (in the magnetic meridian ) be `theta` .IF `B_(H) and B_(Y)` by the horizontal and vertical components of earth.s magnetic field respectively, then `tan theta=(B_(Y)"/"B_(H))` In the plane situated at `30^(@)` with the magnetic meridian the horizontal and vertical components of earth.s magnetic field will be `B_(H)cos30^(@)` while the verticle component will be `B_(Y)` . The angle of dip in the plane is
`tan45^(@)=(B_(Y))/(B_(H)cos30^(@))`
`:.(tantheta)/(tantheta45^(@))=cos30^(@)ortheta=tan^(-1)(sqrt3"/"2)`
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