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Q.

A short magnet oscillates in vibration magneto mater 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 of infinite length placed 20 cm west of the magnet. Then the new frequency is

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a

4  Hz

b

2.5  Hz

c

9  Hz

d

5  Hz

answer is D.

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Detailed Solution

From n=12πMBI As nαB n1n2=Β1B2  B1=BH=12  μT  B2=  resultant of B and BH , Where B=μ0i2π d B=2×10−7×1520×10−2 B=15  μT B2=15−12=3  μT  ∴  10n2=123  ⇒  n2=5  Hz
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