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A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connect to a spring of spring constant $0.5\; Nm^{-1}$ (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance $10\Omega$ and air drag negligible, N will be close to :

Option 1)
10000
Option 2)
1000
Option 3)
5000
Option 4)
50000

Answers (2)

best_answer

$l=10 \; cm$

$k=0.5\; N/m$

$B=0.1\; T$

$-kx-ilB = \frac{md^{2}x}{dt^{2}}.....\; \; \; \; (1)$

$i =\frac{Bl}{R}\frac{dx}{dt} \; \; \; \; \; \; (2)\rightarrow \; Put\; in \left ( 1 \right )$

$-kx-\frac{B^{2}l^{2}}{R}\frac{dx}{dt}=\frac{md^{2}x}{dt^{2}}$

$\Rightarrow A=A_{o}e^{-kt}$

for $A=\frac{A_{o}}{e}=A_{o}e^{-kt}$

$kt=1$

$\Rightarrow \left ( \frac{B^{2}l^{2}}{R.2m} \right )t=1$

$t=\frac{2mR}{B^{2}l^{2}}=\frac{2\times 50\times 10^{-3}\times 10}{0.1\times 0.1\times 10\times 10^{-4}\times 10}$

$t=10^{4}s$

$T_{o}=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{50\times 10^{-3}}{0.5}}=\frac{2\pi }{\sqrt{10}}$

No. of oscillations = $\frac{t}{T_{o}}=\frac{10^{4}\times \sqrt{10}}{2\pi }\approx 5000$

Option 1)

10000

Option 2)

1000

Option 3)
5000
Option 4)

50000

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