If a magnet is suspended at an angle 30^o to | vedclass

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(A) Let the real dip be $\phi$. The vertical component of the Earth's magnetic field is $B_V$ and the horizontal component is $B_H$. The real dip is g

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$	an^{-1}(\sqrt{3}/2)$

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(A) Let the real dip be $\phi$. The vertical component of the Earth's magnetic field is $B_V$ and the horizontal component is $B_H$. The real dip is given by $	an \phi = \frac{B_V}{B_H}$.When the magnet is suspended at an angle $\beta = 30^o$ to the magnetic meridian,the effective horizontal component becomes $B_H' = B_H \cos \beta$.The apparent dip $\phi'$ is given by $	an \phi' = \frac{B_V}{B_H \cos \beta}$.Given $\phi' = 45^o$ and $\beta = 30^o$,we have $	an 45^o = \frac{B_V}{B_H \cos 30^o}$.Since $	an 45^o = 1$ and $\cos 30^o = \frac{\sqrt{3}}{2}$,we get $1 = \frac{B_V}{B_H (\sqrt{3}/2)}$.This simplifies to $\frac{B_V}{B_H} = \frac{\sqrt{3}}{2}$.Therefore,$	an \phi = \frac{\sqrt{3}}{2}$,which implies $\phi = 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.

If a magnet is suspended at an angle $30^o$ to the magnetic meridian,it makes an angle of $45^o$ with the horizontal. The real dip is

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