Question

The real angle of dip at a place, if a magnet is suspended at an angle of \(30^\circ\) to the magnetic meridian and the dip needle makes an angle of \(45^\circ\) with the horizontal, is:

• A. \(\tan^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)\)
• B. \(\tan^{-1}(\sqrt{3})\)
• C. \(\tan^{-1}\!\left(\sqrt{\dfrac{3}{2}}\right)\)
• D. \(\tan^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right)\)

Hint:

If the dip needle is not in the magnetic meridian: \[ \tan(\text{apparent dip}) = \frac{\tan(\text{real dip})}{\cos \theta} \] Always check whether the needle is aligned with the magnetic meridian.

Answer 1

The Correct Option is A

Step 1: Identify the given quantities. Angle between magnet and magnetic meridian: \[ \theta = 30^\circ \] Apparent angle of dip: \[ \delta' = 45^\circ \]
Step 2: Recall the relation between real dip and apparent dip. When the magnet is not in the magnetic meridian, the relation is: \[ \tan \delta' = \frac{\tan \delta}{\cos \theta} \] where \(\delta\) = real angle of dip.
Step 3: Substitute the given values. \[ \tan 45^\circ = \frac{\tan \delta}{\cos 30^\circ} \] \[ 1 = \frac{\tan \delta}{\frac{\sqrt{3}}{2}} \]
Step 4: Solve for the real angle of dip. \[ \tan \delta = \frac{\sqrt{3}}{2} \] \[ \delta = \tan^{-1}\!\left(\frac{\sqrt{3}}{2}\right) \]
Hence, the real angle of dip is \(\boxed{\tan^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)}\).