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

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Answer 1

The Correct Option is D

The problem involves determining the new time period of oscillation for a magnet in a vibration magnetometer, given a change in the magnetic field. Let's solve it step-by-step: 

1. Initially, the magnet has a time period \(T_1 = 2\) seconds in the Earth's horizontal magnetic field \(B_1 = 24\ \mu T\).
2. The formula for the time period of oscillation of a magnet in a magnetic field is given by: \(T = 2\pi \sqrt{\frac{I}{mB}}\),
3. where:
   • \(I\) is the moment of inertia of the magnet.
   • \(m\) is the magnetic moment.
   • \(B\) is the magnetic field strength.
4. When a horizontal magnetic field of \(18\ \mu T\) is applied in a direction opposite to the Earth's magnetic field, the resultant magnetic field strength \(B_2\) is: 
   \(B_2 = B_1 - 18\ \mu T = 24\ \mu T - 18\ \mu T = 6\ \mu T\).
5. Let \(T_2\) be the new time period, then: 
   \(T_2 = 2\pi\sqrt{\frac{I}{mB_2}}\).
6. Using the relation for time periods: 
   \(\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}\),
7. we can substitute values to find \(T_2\): 
   \(\frac{T_2}{2} = \sqrt{\frac{24}{6}}\), 
   \(\frac{T_2}{2} = \sqrt{4} = 2\), 
   \(T_2 = 2 \times 2 = 4\) seconds.
8. Thus, the new time period of the magnet is \(4\) seconds.

Therefore, the correct answer is: 4 s