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:

Options

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

Answer

A

Explanation

let `theta =` dip angle
H = horizontal component of the Earth's field
V = vertical compnent of the Earth's field
Also. `tantheta = (V)/(H)`
but `tan45^(@) = (V)/(Hcos30^(@))`
`therefore (tan theta)/(tan 45^(@)) = cos 30^(@)`
or `theta = tan^(-1) ((sqrt(3))/(2))`