QJEEM19Ph015d: JEE Main 2019 Physics - Class 12 JEEM19Ph015d Solution: A thin strip 10 cm long is on a U shaped wire of negligible…

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

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