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A magnet makes an angle of 45° with the ...

A magnet makes an angle of 45° with the horizontal in a plane making an angle of `30^@` with the magnetic meridian. The true value of the dip angle at the place is .....

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Suppose the true value of the dip angle at the, given place is `theta_(1)`.
`therefore tan phi = (B_v)/( B_H)" "…(1)`
Here a magnet makes an angle of 45° with the horizontal in a plane making an angle 30° with the magnetic meridian. At this place, horizontal component of magnetic field of the earth will be `B_H cos 30^@` and vertical component will remain `B_(v)`.
`therefore tan 45^(@) (B_v)/( B_H cos 30^(@) ) " "...(2)`
Taking ratio of equation (1) and (2),
`therefore (tan phi)/( tan 45^(@) ) = ((B_v)/(B_H))/( (B_v)/( B_H cos 30^@) )`
`therefore (tan phi )/( 1) = cos 30^@`.
`therefore tan phi = (sqrt(3) )/( 2) `
`therefore tan phi = (1.732)/( 2)`
`therefore phi = tan^(-1) (0.866)`
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