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) \[90{}^\circ \]                                   

B) \[60{}^\circ \]                   

C) \[45{}^\circ \]                                                   

D) \[30{}^\circ \]

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, \[T=3\sqrt{2}s,T=3s\]                 \[\therefore \]  \[\frac{T}{T}=\sqrt{\frac{H}{V}}=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}\] 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                 \[\therefore \]  \[\sqrt{\frac{H}{V}}=\sqrt{\cos \phi }=\frac{1}{\sqrt{2}}\]                 \[\therefore \]  \[\cos \phi \text{=}\frac{1}{2}\]                 \[\Rightarrow \]               \[\phi =\,\text{6}{{\text{0}}^{o}}\]