There are two forces on slider.

Spring force = kx

where, k = spring constant.

As the slider is kept in a uniform magnetic field B = 0.1 T, hence it will experience a force, i.e.

Magnetic force = Bil

where, l = length of the strip.

Now, using

F_net = ma

We have,

$$( - kx) + ( - Bil) = ma$$

$$ \Rightarrow - kx - Bil - ma = 0$$

$$ \Rightarrow - kx - {{{B^2}{l^2}} \over R}.\,v - ma = 0$$ [$$\because$$ $$i = {{Blv} \over R}$$]

and acceleration, $$a = {{{d^2}x} \over {d{t^2}}}$$

Hence, the modified equation becomes

$$ \Rightarrow {{m{d^2}x} \over {d{t^2}}} + {{{B^2}{l^2}} \over R}\left( {{{dx} \over {dt}}} \right) + kx = 0$$

This is the equation of damped simple harmonic motion.

So, amplitude of oscillation varies with time as

$$A = {A_e}^{ - {{{B^2}{l^2}} \over {2Rm}}.\,t}$$

Now, when amplitude is $${{{A_0}} \over e}$$, then

$${{{A_0}} \over e} = {{{A_0}} \over {{e^{{{{B^2}{l^2}} \over {2Rm}}.\,t}}}}$$ (as given)

$$ \Rightarrow \left( {{{{B^2}{l^2}} \over {2Rm}}} \right)t = 1$$ or $$t = {{2Rm} \over {{B^2}{l^2}}}$$

According to the question, magnetic field B = 0.1 T, mass of strip m = 50 $$\times$$ 10^$$-$$3 kg, resistance R = 10 $$\Omega$$, l = 10 cm = 10 $$\times$$ 10^$$-$$2 m

$$\therefore$$ $$t = {{2Rm} \over {{B^2}{l^2}}} = {{2 \times 10 \times 50 \times {{10}^{ - 3}}} \over {{{(0.1)}^2} \times {{(10 \times {{10}^{ - 2}})}^2}}}$$

$$ = {1 \over {{{10}^{ - 4}}}} = 10000$$ s

Given, spring constant, k = 0.5 Nm^$$-$$1

Also, time period of oscillation is

$$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{{50 \times {{10}^{ - 3}}} \over {0.5}}} = {{2\pi } \over {\sqrt {10} }} \approx 2s$$

So, number of oscillations is

$$N = {t \over T} = {{10000} \over 2} = 5000$$