A thin strip $10\ \text{cm}$ long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant $0.5\ \text{N m}^{-1}$ (see figure). The assembly is kept in a uniform magnetic field of $0.1\ \text{T}$. 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
Answer: (B) $5000$
At speed $v$ the strip has emf $Blv$ and current $Blv/R$, so the magnetic force opposing the motion is
$$F = \frac{B^2l^2}{R}v = bv,\quad b = \frac{(0.1)^2(0.1)^2}{10} = 10^{-5}\ \text{kg/s}$$
For damped oscillations the amplitude falls as $e^{-bt/2m}$, so it drops by a factor $e$ after
$$t = \frac{2m}{b} = \frac{2\times0.05}{10^{-5}} = 10^4\ \text{s}$$
The period is $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.1} \approx 1.99\ \text{s}$, so
$$N = \frac tT \approx 5000$$
Solution by Sreeraj P, M.Sc Physics