Q. 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
A) 4 s
B) 1s
C) 2 s
D) 3s
Explanation
[Solution from source] T ∝ 1/√B. The opposing field reduces the net horizontal field to 24 − 18 = 6 μT. T′ = 2 × √(24/6) = 4 s. Adding the fields (42 μT) would give a shorter period. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.