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Q.

A short magnet oscillates in vibration magnetometer with a frequency 10 Hz where horizontal component of earth's magnetic field is 12 μT. A downward current of 15 A is established in the vertical wire placed 20 cm west of the magnet. New frequency is

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a

4Hz

b

2.5Hz

c

9Hz

d

5Hz

answer is D.

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Detailed Solution

B1=μ0i2πr=150 x 10-7=15 μTB=B1-BH=15-3 μT=12 μTf1f=BHB=3 μT12 μT=12 ; f1=102=5 Hz
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