A short magnet oscillates with a time period | vedclass

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(C) Given: $T_1 = 0.1 \, s$, $B_H = 24 \, \mu T = 24 	imes 10^{-6} \, T$, $I = 18 \, A$, $r = 20 \, cm = 0.2 \, m$.The magnetic field $B$ due to the

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$0.076$

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(C) Given: $T_1 = 0.1 \, s$, $B_H = 24 \, \mu T = 24 	imes 10^{-6} \, T$, $I = 18 \, A$, $r = 20 \, cm = 0.2 \, m$.The magnetic field $B$ due to the vertical wire at the position of the magnet is given by $B = \frac{\mu_0 I}{2 \pi r}$.$B = \frac{4 \pi 	imes 10^{-7} 	imes 18}{2 \pi 	imes 0.2} = 18 	imes 10^{-6} \, T = 18 \, \mu T$.Since the wire is to the east and the current is downward, by the right-hand rule, the magnetic field $B$ produced by the wire at the magnet's location points towards the north (same direction as $B_H$).The resultant magnetic field is $B_{net} = B_H + B = 24 \, \mu T + 18 \, \mu T = 42 \, \mu T$.The time period of oscillation is $T = 2 \pi \sqrt{\frac{I}{MB}}$, so $T \propto \frac{1}{\sqrt{B}}$.Therefore, $\frac{T_2}{T_1} = \sqrt{\frac{B_H}{B_{net}}} = \sqrt{\frac{24}{42}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.$T_2 = T_1 	imes \frac{2}{\sqrt{7}} = 0.1 	imes \frac{2}{2.645} \approx 0.076 \, s$.

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