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 –
Answers
The FGM-301 is the best product for measuring DC and low frequency magnetic fields in three dimensions. This is a compact magnetic sensing device suitable for a wide variety of magnetic field measurement applications. Being an analog sensor, the FGM-301 provides high resolution and low signal delay. This product is far more stable and linear than magneto-resistive and magneto-inductive magnetometers.
Watson Industries has been developing state-of-the-art fluxgate magnetometers for almost thirty years to provide the FGM-301. As a result, this magnetometer is relied on for precision measurements in very critical projects around the world, including near the north and south poles.
Explanation:
The right answer of the given question is 4s. The time period of a vibration magnetometer will be
T 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁ √(B₁ / B₂)
2 √ (24 x 10⁻⁶ / 6 x 10⁻⁶) = 4 sec.
Further, you can discuss with your teacher or in your friend circle to have a better understanding of the question.