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

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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 2

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.