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A magnet suspended at 30^@ with magnetic...

A magnet suspended at `30^@` with magnetic meridian makes an angle of `45^@` with the horizontal. What shall be the actual value of the angle of dip?

A

`tan^(-1)(sqrt(3)/(2))`

B

`tan^(-1) (sqrt(3))`

C

`45^@`

D

`30^@`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • If a dip needle is suspended at an angle 30° to the magnetic meridian, it makes an angle of 45^@ with the horizontal. The real dip is

    A
    `tan^(-1)sqrt(3)/2`
    B
    `tan ^(-1)sqrt(3)`
    C
    `tan^(-1)sqrt(3/2)`
    D
    `tan^(-1)2/sqrt(3)`

  • 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)`

  • The real angle of dip, if a magnet is suspended at an angle of 30^(@) to the magnetic meridian and the dip needle makes an angle of 45^(@) with horizontal, is:

    A
    `tan^(-1)(sqrt(3)/2)`
    B
    `tan^(-1)(sqrt(3))`
    C
    `tan^(-1)(sqrt(3)/sqrt(2))`
    D
    `tan^(-1)(2/sqrt(3))`

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    A freely suspended bar magnet makes an angle of 60^@ with the horizontal component and 45^@ with the magnetic meridian of a place. Calculate the angle of dip for that place.

    If a magnet is suspended at an angle 30^(@) to the magnetic meridian and in this plane the dip needle makes an angle of 60^(@) with the horizontal. The true value of dip is:

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