A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be
- 1 s
- 2 s
- 3 s
- 4 s
Answer
(4) 4 s
Solution
The time period of a magnet oscillating in a field B is T = 2 ImB, so T 1 B.
The net field now is B' = 24 - 18 = 6 , hence
T'T = BB' = 246 = 2
T' = 2 2 = 4 s
Answer: (4)