Q. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec 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) 1$s$
C) 2$s$
D) $3s$
Explanation
(a) $T = 2\pi\sqrt{\frac{1}{MB_{H}}},\ T' = 2\pi\sqrt{\frac{1}{M(B_{H} - B)}} \Rightarrow T^{'} = 2T = 4s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.