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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4 Hz

2.5 Hz

9 Hz

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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