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

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4 Hz

9 Hz

5 Hz

2.5 Hz

answer is D.

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Detailed Solution

$B' = B_{H} - B = \frac{μ_{o} I}{2 πa} - B_{H}$

$\Rightarrow B' = \frac{10^{- 7} \times 2 \times 15}{20 \times 10^{- 2}} - (12 \times 10^{- 6}) = (15 - 12) \times 10^{- 6} T = 3 μT$

$Now, \frac{f'}{f} = \sqrt{\frac{B'}{B}} = \sqrt{\frac{3}{12}} = \frac{1}{2}$

$\Rightarrow f' = \frac{10}{2} = 5 Hz$

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