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A short magnet oscillates in vibration magneto mater with a frequency $10 H z$ where horizontal component of earth’s magnetic field is 12 $μ T .$ 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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$4 H z$

$2.5 H z$

$9 H z$

$5 H z$

answer is D.

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Detailed Solution

From $n = \frac{1}{2 π} \sqrt{\frac{M B}{I}}$ As $n α \sqrt{B}$

$\frac{n_{1}}{n_{2}} = \sqrt{\frac{Β_{1}}{B_{2}}}$ $B_{1} = B_{H} = 12 μ T$

$B_{2} =$ resultant of B and $B_{H}$ , Where $B = \frac{μ_{0} i}{2 π d}$

$B = \frac{2 \times 10^{- 7} \times 15}{20 \times 10^{- 2}}$ $B = 15 μ T$

$B_{2} = 15 - 12 = 3 μ T$ $∴ \frac{10}{n_{2}} = \sqrt{\frac{12}{3}} \Rightarrow n_{2} = 5 H z$

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