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 \[\therefore \] \[\sqrt{\frac{H}{V}}=\sqrt{\cos \phi }=\frac{1}{\sqrt{2}}\] \[\therefore \] \[\cos \phi \text{=}\frac{1}{2}\] \[\Rightarrow \] \[\phi =\,\text{6}{{\text{0}}^{o}}\]You need to login to perform this action.
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