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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
Options
(a) 1 s
(b) 2 s
(c) 3 s
(d) 4 s
Correct Answer:
4 s
Explanation:
Time period of a vibration magnetometer,
T 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁ √(B₁ / B₂)
2 √ (24 x 10⁻⁶ / 6 x 10⁻⁶) = 4 sec.
Related Questions: - A stone of mass 1kg is tied to a string 4m long and is rotated at constant speed
- A paramagnetic sample shows a net magnetisation of 0.8 Am⁻¹, when placed
- A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide
- Vectors A⃗, B⃗ and C⃗ are such that A⃗ . B⃗ = 0 and A⃗ . C⃗ = 0 Then the vector parallel
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A stone of mass 1kg is tied to a string 4m long and is rotated at constant speed
- A paramagnetic sample shows a net magnetisation of 0.8 Am⁻¹, when placed
- A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide
- Vectors A⃗, B⃗ and C⃗ are such that A⃗ . B⃗ = 0 and A⃗ . C⃗ = 0 Then the vector parallel
- A tuning fork of frequency x produces 4 beats with a source of 256 Hz and 8 beats
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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