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Step-by-Step Solution
Step 1: Identify the Physical Situation
A thin strip of length $l = 10\,\text{cm}$ is placed on a U-shaped wire. It is connected to a spring of spring constant $k=0.5\,\text{N/m}$. The system is in a uniform magnetic field $B = 0.1\,\text{T}$, and the strip has a mass $m = 50\,\text{g}$ (i.e., $m = 0.05\,\text{kg}$) and resistance $R = 10\,\Omega$. The strip can oscillate when displaced from its equilibrium position.
Step 2: Identify the Forces Acting on the Strip
1. Spring Force: This is $F_\text{spring} = -kx$, where $x$ is the displacement from the equilibrium.
2. Magnetic Damping Force: As the strip moves in the magnetic field, a current is induced. The force due to the induced current is $F_\text{mag} = -Bil$, where $i$ is the induced current and $l$ is the length of the strip.
Step 3: Express the Induced Current
When the strip moves with velocity $v = \frac{dx}{dt}$ in the magnetic field, an emf $e = Blv$ is induced. Using Ohm’s law, the current is
$ i = \frac{e}{R} = \frac{Bl\,\frac{dx}{dt}}{R}. $
Hence, the magnetic force can be written as
$ F_\text{mag} = B \, i \, l = B \left(\frac{Bl\,\frac{dx}{dt}}{R}\right) l
= \frac{B^2l^2}{R} \,\frac{dx}{dt}. $
Step 4: Write the Equation of Motion with Damping
Newton’s second law gives
$ m \frac{d^2 x}{dt^2} = -kx - \frac{B^2 l^2}{R} \frac{dx}{dt}. $
This is the standard form of a damped harmonic oscillator:
$ m \frac{d^2 x}{dt^2} + \left(\frac{B^2 l^2}{R}\right)\frac{dx}{dt} + kx = 0. $
Step 5: Determine the Damping Constant and Decay of Amplitude
For a damped simple harmonic motion, the amplitude $A(t)$ decays over time according to:
$ A(t) = A_0 \, e^{-\frac{\gamma}{2} t}, $
where $\gamma$ is the damping coefficient. From the equation of motion, we identify
$ \gamma = \frac{B^2 l^2}{R m}. $
When the amplitude drops by a factor of $e$, we have:
$ A(t) = \frac{A_0}{e} = A_0 \, e^{-\frac{\gamma}{2} t}. $
Taking natural logarithms and simplifying,
$ e^{-\frac{\gamma}{2} t} = \frac{1}{e} \quad \Rightarrow \quad \frac{\gamma}{2} t = 1 \quad \Rightarrow \quad t = \frac{2}{\gamma}. $
Thus,
$ t = \frac{2}{\gamma} = \frac{2Rm}{B^2l^2}. $
Step 6: Calculate the Time $t$ for Amplitude to Decay by $e$
Plugging in the values:
$ R = 10\,\Omega, \quad m = 0.05\,\text{kg}, \quad B = 0.1\,\text{T}, \quad l = 0.10\,\text{m}. $
So,
$ t = \frac{2 \times 10 \times 0.05}{(0.1)^2 \times (0.10)^2}
= \frac{2 \times 10 \times 0.05}{0.01 \times 0.01}
= \frac{1}{10^{-4}}
= 10000\,\text{s}. $
Step 7: Determine the Time Period of the Undamped Oscillation
For a mass-spring system (ignoring damping for the moment), the time period is
$ T = 2\pi \sqrt{\frac{m}{k}}. $
Substitute $m = 0.05\,\text{kg}$ and $k = 0.5\,\text{N/m}$:
$ T = 2\pi \sqrt{\frac{0.05}{0.5}}
= 2\pi \sqrt{\frac{1}{10}}
= \frac{2\pi}{\sqrt{10}}
\approx 2\,\text{s}. $
Step 8: Calculate the Number of Oscillations Before Amplitude Reduces by $e$
In the time interval $t = 10000\,\text{s}$, each oscillation takes about $T \approx 2\,\text{s}$. Therefore, the number of oscillations $N$ is
$ N = \frac{t}{T} = \frac{10000}{2} = 5000. $
Final Answer
The strip performs approximately 5000 oscillations before its amplitude decreases by a factor of $e$.
