A thin strip 10, cm long is on a U shaped wir | vedclass

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(D) The motion of the strip is a damped harmonic oscillation due to the magnetic force (Lorentz force) acting as a damping force.The damping force is

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$5000$

Vedclass · Answer

(D) The motion of the strip is a damped harmonic oscillation due to the magnetic force (Lorentz force) acting as a damping force.The damping force is $F_d = -B \ell I = -B \ell \left( \frac{B \ell v}{R} ight) = -\frac{B^2 \ell^2}{R} v$.Comparing this with the damping force equation $F_d = -bv$, we get the damping constant $b = \frac{B^2 \ell^2}{R}$.Given: $B = 0.1\, T$, $\ell = 0.1\, m$, $R = 10\, \Omega$, $m = 0.05\, kg$, $k = 0.5\, N/m$.$b = \frac{(0.1)^2 	imes (0.1)^2}{10} = \frac{10^{-2} 	imes 10^{-2}}{10} = 10^{-5}\, kg/s$.The time constant for the decay of amplitude is $	au = \frac{2m}{b} = \frac{2 	imes 0.05}{10^{-5}} = \frac{0.1}{10^{-5}} = 10^4\, s$.After time $t = 	au$, the amplitude becomes $A = A_0 e^{-1}$.The number of oscillations $N = \frac{t}{T_0}$, where $T_0 = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.05}{0.5}} = 2\pi \sqrt{0.1} = \frac{2\pi}{\sqrt{10}}$.$N = \frac{10^4}{2\pi / \sqrt{10}} = \frac{10^4 	imes \sqrt{10}}{2\pi} \approx \frac{10000 	imes 3.162}{6.28} \approx 5035$.Thus, $N$ is close to $5000$.

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