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 \[\therefore \] \[\sqrt{\frac{H}{V}}=\sqrt{\cos \phi }=\frac{1}{\sqrt{2}}\] \[\therefore \] \[\cos \phi =\frac{1}{2}\] \[\Rightarrow \] \[\phi ={{60}^{o}}\]You need to login to perform this action.
You will be redirected in
3 sec