A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizon

Question

A magnet hung at $45^{\circ}$ with magnetic meridian makes an angle of $60^{\circ}$ with the horizontal. The actual value of the angle of dip is\

Professor PC Thomas Classes · Answer

(a) : Let $B$ be the magnetic field at some point. $H$ and $V$ be the horizontal and vertical components of $B$ and $	heta$ is the actual angle of dip at that point.\ We know that, $H=B \cos 	heta$ and $V=B \sin 	heta$ $$\begin{equation*} 	herefore 	an 	heta=\frac{V}{H} 	ag{i} \end{equation*}$$ In vertical plane at $45^{\circ}$ from the magnetic meridian, horizontal component is\ $H^{\prime}=H \cos 45^{\circ}=\frac{H}{\sqrt{2}}$\ Vertical component remains the same i.e., $V$. Therefore, apparent dip will be given as, $$\begin{equation*} 	an 	heta^{\prime}=\frac{V}{H^{\prime}}=\frac{V}{H / \sqrt{2}}=\frac{\sqrt{2} V}{H} 	ag{ii} \end{equation*}$$ Dividing equation (i) by (ii), we have\ $\frac{	an 	heta}{	an 	heta^{\prime}}=\frac{V}{H} 	imes \frac{H}{\sqrt{2} V}=\frac{1}{\sqrt{2}}$\ or $	an 	heta=\frac{1}{\sqrt{2}} 	an 	heta^{\prime}=\frac{1}{\sqrt{2}} 	an 60^{\circ}$\ ( $\because 	heta^{\prime}=60^{\circ}$ (given))\ $=\frac{1}{\sqrt{2}} 	imes \sqrt{3}=\sqrt{\frac{3}{2}} \quad 	herefore \quad 	heta=	an ^{-1}\left(\sqrt{\frac{3}{2}}ight)$\