Question 210: Magnetism and Matter

Question

A thin strip 10 cm long is on a U shaped wire
of negligible resistance and it is connected 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 oscillation it performs before its
amplitude decreases by a factor of e is N. If the
mass of the strip is 50 grams, its resistance 10W
and air drag negligible, N will be close to :
![](https://imagex.cdn.examgoal.net/GooBZSyp6SXvTaS8A/NVmggyM29ilqbnA0UeKdpKKdq256R/iKz8bWzxxzXWpz64UUG1Su/image.png)

Accepted Answer

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}}} ight) + 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}}} ight)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 $	imes$ 10
$-$3
kg, resistance R = 10 $\Omega$, l = 10 cm = 10 $	imes$ 10
$-$2
m
$	herefore$ $t = {{2Rm} \over {{B^2}{l^2}}} = {{2 	imes 10 	imes 50 	imes {{10}^{ - 3}}} \over {{{(0.1)}^2} 	imes {{(10 	imes {{10}^{ - 2}})}^2}}}$
$ = {1 \over {{{10}^{ - 4}}}} = 10000$ s
Given, spring constant, k = 0.5 Nm
$-$1
Also, time period of oscillation is
$T = 2\pi \sqrt {{m \over k}}  = 2\pi \sqrt {{{50 	imes {{10}^{ - 3}}} \over {0.5}}}  = {{2\pi } \over {\sqrt {10} }} \approx 2s$
So, number of oscillations is
$N = {t \over T} = {{10000} \over 2} = 5000$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l36rep7l/ea20cd35-5aa5-4de5-8307-038452e786c0/2747e910-d402-11ec-b808-5752a3163b13/file-1l36rep7m.png?format=png)

Answer

50000

Answer

1000

Answer

5000

Answer

10000

Magnetism and Matter

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