NEET-XII-Physics
36: Permanent Magnets
- #14A magnetic needle is free to rotate in a vertical plane which makes an angle of 60° with the magnetic meridian. If the needle stays in a direction making an angle of
tan-1 23with the horizontal, what would be the dip at that place?Ans : Given:
Angle made by the magnetic meridian with the plane of rotation of the needle, `` \theta =60°``
Angle made by the needle with the horizontal, `` {\delta }_{1}={\,\mathrm{\,tan\,}}^{-1}\left(\frac{2}{\sqrt{3}}\right)``
If `` \delta `` is the angle of dip, then
`` \,\mathrm{\,tan\,}{\delta }_{1}=\frac{\,\mathrm{\,tan\,}\delta }{\,\mathrm{\,cos\,}\theta }``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\,\mathrm{\,tan\,}{\delta }_{1}\,\mathrm{\,cos\,}\theta ``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\,\mathrm{\,tan\,}\left({\,\mathrm{\,tan\,}}^{-1}\frac{2}{\sqrt{3}}\right)\,\mathrm{\,cos\,}60°``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}``
`` \Rightarrow \delta =30°``
Page No 278: