Question 201: 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 :

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

Answer

50000

Answer

1000

Answer

5000

Answer

10000

Magnetism and Matter

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