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(C) The time period of an oscillating magnet is given by $T = 2\pi \sqrt{\frac{I}{m B_H}}$,where $I$ is the moment of inertia,$m$ is the magnetic mome

Vedclass · Accepted Answer

$0.076$

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(C) The time period of an oscillating magnet is given by $T = 2\pi \sqrt{\frac{I}{m B_H}}$,where $I$ is the moment of inertia,$m$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.Initially,$T = 0.1 \, s$ and $B_H = 24 \, \mu T$.The magnetic field $B$ produced by a long vertical wire carrying current $i$ at a distance $a$ is $B = \frac{\mu_0}{4\pi} \cdot \frac{2i}{a}$.Given $i = 18 \, A$ and $a = 20 \, cm = 0.2 \, m$,we have $B = 10^{-7} \cdot \frac{2 	imes 18}{0.2} = 18 	imes 10^{-6} \, T = 18 \, \mu T$.Since the wire is to the east and the current is downward,by the right-hand rule,the magnetic field $B$ at the magnet's position points north,in the same direction as $B_H$.Thus,the new total magnetic field is $B_{net} = B_H + B = 24 \, \mu T + 18 \, \mu T = 42 \, \mu T$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I}{m(B_H + B)}}$.Taking the ratio: $\frac{T'}{T} = \sqrt{\frac{B_H}{B_H + B}} = \sqrt{\frac{24}{24 + 18}} = \sqrt{\frac{24}{42}} = \sqrt{\frac{4}{7}} \approx 0.7559$.Therefore,$T' = 0.1 	imes 0.7559 \approx 0.0756 \, s \approx 0.076 \, s$.

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

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