In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is 2 sec. When the magnet is brought near a field magnet F and parallel to it, the time period reduces to 1 sec. The ratio of F/H of the horizontal component H and the field F due to magnet will be
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a
3
b
1/3
c
3
d
13
answer is A.
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Detailed Solution
We know that T=2πIMBH Here, T=2=2πIMH1/2 where BH=HWhen &e magnet is brought near and parallel to it, the time period reduces to 1 sec. i.e., time period decreases. Hence the field is now (H + F). HenceT′=1=2πIM(H+F)1/2∴21=H+FH1/2 or 41=H+FH or 4=1+FH or FH=3