Question
Question: A bar magnet of moment of inertia \[9\times {{10}^{-3}}kg{{m}^{2}}\] placed in a vibration magnetome...
A bar magnet of moment of inertia 9×10−3kgm2 placed in a vibration magnetometer and oscillating in a uniform magnetic field 16π2×10−5T makes 20 oscillations in 15 s. The magnetic moment of the bar magnet is
& A.300A{{m}^{2}} \\\ & B.200A{{m}^{2}} \\\ & C.500A{{m}^{2}} \\\ & D.600A{{m}^{2}} \\\ & E.400A{{m}^{2}} \\\ \end{aligned}$$Solution
Hint: If there are number of oscillations in a given time period, then we have to find time period for one oscillation and apply the formula of time period for a bar magnet vibrating in given magnetic field that is given by T=2πmBI.
Formula Used:
T=2πmBI
Complete step by step answer:
In our question, we are given with:
Moment of inertia of bar magnet =9×10−3kgm2
Magnetic Field =16π2×10−5T
Time period for 20 oscillations = 15 second
And we have to find the magnetic moment of the bar magnet.
Firstly we will find the time period for one oscillation.
20 oscillations = 15 seconds
1 oscillation = 2015 seconds
Time Period of oscillation due magnetic field is given by:
T=2πmBI
Where:
T=$$$$\dfrac{15}{20} seconds
I=moment of inertia
m=magnetic moment of bar magnet
B=given magnetic field