Magnetics Ques 136
- A thin strip $10$ $ cm$ long is on an $U$-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 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
(2019 Main, 8 April I)
(a) $1000$
(b) $50000$
(c) $5000$
(d) $10000$
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Answer:
Correct Answer: 136.(c)
Solution:
Formula:
Magnetic Force Acting On A Current Carrying Wire:
- There are two forces on slider
Spring force $=k x$
where, $\quad k=$ 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, $\quad l=$ length of the strip.
Now, using $F _{\text {net }}=m a$
We have,
$ (-k x)+(-B i l)=m a $
$ \begin{array}{ll} \Rightarrow & -k x-B i l-m a=0 \\ \Rightarrow & -k x-\frac{B^{2} l^{2}}{R} \cdot v-m a=0 \quad [\because i=\frac{B l v}{R}] \end{array} $
and acceleration, $a=\frac{d^{2} x}{d t^{2}}$
Hence, the modified equation becomes
$ \Rightarrow \quad \frac{m d^{2} x}{d t^{2}}+\frac{B^{2} l^{2}}{R} (\frac{d x}{d t})+k x=0 $
This is the equation of damped simple harmonic motion.
So, amplitude of oscillation varies with time as
$ A=A _0 e^{-\frac{B^{2} l^{2}}{2 R m} \cdot t} $
Now, when amplitude is $\frac{A _0}{e}$, then
$ \begin{aligned} \frac{A _0}{e} & =\frac{A _0}{e^{\frac{B^{2} l^{2}}{2 R m} \cdot t}} \\ \Rightarrow \quad (\frac{B^{2} l^{2}}{2 R m}) t & =1 \text { or } t=\frac{2 R m}{B^{2} l^{2}} \end{aligned} $
According to the question, magnetic field $B=0.1 T$, mass of strip $m=50 \times 10^{-3} kg$,
resistance $R=10 \Omega, l=10 cm=10 \times 10^{-2} m$
$ \begin{aligned} \therefore \quad t=\frac{2 R m}{B^{2} l^{2}} & =\frac{2 \times 10 \times 50 \times 10^{-3}}{(0.1)^{2} \times\left(10 \times 10^{-2}\right)^{2}} \\ & =\frac{1}{10^{-4}}=10000 s \end{aligned} $
Given, spring constant, $k=0.5 Nm^{-1}$
Also, time period of oscillation is
$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{50 \times 10^{-3}}{0.5}}=\frac{2 \pi}{\sqrt{10}} \approx 2 s $
So, number of oscillations is $N=\frac{t}{T}=\frac{10000}{2}=5000$
BETA
