A vibration magnetometer placed in magnetic meridian has a small bar magnet

A Vibration Magnetometer Placed In Magnetic Meridian Has A Small Physics Question

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:

  1. Characteristic X-rays are produced due to
  2. An alternating voltage given as, V=100√2 sin100t V is applied to a capacitor
  3. A magnetic field of 1600 A/m produces a magnetic flux of 2.4 x 10⁻⁵ Wb
  4. A thermally insulated rigid container contains an ideal gas heated by a filament
  5. A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken

Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*