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
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0. 2. 4 d. M π μ. (4). 3. 0. 22. 4 d. M π μ. 6. Figure shows two identical short magnetic dipoles a and b of .... magnetometer and are allowed to oscillate in the earth's ... A small bar magnet having a magnetic moment of ... a current of 10 A. Find the time period of oscillation ... of horizontal component of earth's magnetic field .
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Answer:
d) 4s
Explanation:
The time period T of oscillation of a magnet is given as -
T = 2π√I/MB where,
l = Moment of inertia of the magnet about the axis of rotation
M = Magnetic moment of the magnet
B = Uniform magnetic field As l, B remains the same
Therefore,
T∝1/√B or T2/T1 = √B1/B2
= B1 = 24ut
B2 = 24ut - 18ut
B2 = 6ut
T1 = 2 seconds
Hence, T2 = (2)√ 24ut/6ut
= 4
Thus, 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 4 seconds.