question_answer

The time of vibration of a dip needle vibration in the vertical plane in the magnetic meridian is 3 s. When the same magnetic needle is made to vibrate in the horizontal plane, the time of vibration is \[3\sqrt{2}\]s. Then angle of dip will be

A)  \[{{90}^{o}}\]

B)  \[{{60}^{o}}\]

C)  \[{{45}^{o}}\]

D)  \[{{30}^{o}}\]

Correct Answer: B

Solution :

Key Idea: In vertical plane in magnetic meridian both horizontal and vertical components of magnetic field exist. When M is magnetic moment of the magnet, H and V are the horizontal and vertical components of earths magnetic field and I is moment of inertia of magnet about its axis of vibration, then the time-period of-magnet is                                 \[T=2\pi \sqrt{\frac{I}{MH}}\] when horizontal component is taken                 \[T=2\pi \sqrt{\frac{I}{MB}}\] [as in vertical plane in magnetic meridian both V and H act on the needle] Given, \[T=3\sqrt{2}\,s,\,T=3\,s\] \[\therefore \]  \[\frac{T}{T}=\sqrt{\frac{H}{V}}=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}\] Also the angle of dip at a place is the angle between the direction of earths magnetic field and the horizontal in the magnetic meridian at that place \[\therefore \]  \[\sqrt{\frac{H}{V}}=\sqrt{\cos \phi }=\frac{1}{\sqrt{2}}\] \[\therefore \]  \[\cos \phi =\frac{1}{2}\] \[\Rightarrow \] \[\phi ={{60}^{o}}\]