Question

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. $30{}^\circ$

Answer 1

Answer: (B)

$60{}^\circ$

Answer 2

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 earth's magnetic field and $I$ is moment of inertia of magnet about its axis of vibration, then the time-period of magnet is $T=2 \pi \sqrt{\frac{I}{M H}}$ when horizontal component is taken $T^{'}=2 \pi \sqrt{\frac{I}{M B}}$ [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 earth's magnetic field and the horizontal in the magnetic meridian at that place $\therefore \sqrt{\frac{H}{V}}=\sqrt{\cos \phi}=\frac{1}{\sqrt{2}}$ $\therefore \cos \phi=\frac{1}{2}$ $\Rightarrow \phi=60^{\circ}$