Permanent magnetismHard
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 magentic field of 24 microtesla A 18 microtesla magnetic field is produced opposite to the earth′s field by placing a current carrying wire, the new time period of magnet will be
Options
A.1s
B.2s
C.3s
D.4s
Solution
The timpe period T of oscillation of magnet is given by
T = 2π
where,
I = Moment of inertia of the magnet about the axis of rotation
M = Magnetic moment of the magnet
B = Uniform magnetic field
As the I, B remains the same
∴ T ∝
According to given problem,
B1 = 24 μT
B2 = 24 μT - 18 μT = 6 μT
T1 = 2s
∴ T2 = (2s)
= 4s.
T = 2π
where,
I = Moment of inertia of the magnet about the axis of rotation
M = Magnetic moment of the magnet
B = Uniform magnetic field
As the I, B remains the same
∴ T ∝
According to given problem,
B1 = 24 μT
B2 = 24 μT - 18 μT = 6 μT
T1 = 2s
∴ T2 = (2s)
= 4s.Create a free account to view solution
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