A thin strip 10 cm long is on an U-shaped wire of negligible resistance and it is connected to a spr

Q: Why does the magnetic force act in the opposite direction to the velocity?

This is a direct consequence of Lenz's Law. The induced current flows in a direction that creates a magnetic field opposing the change in magnetic flux. Macroscopically, this manifests as a Lorentz force that always opposes the velocity of the conductor, acting as a damping mechanism.

Q: Why can we use the standard damped SHM formula here?

The magnetic force we derived is $F_m = -\frac{B^2l^2}{R}v$. Notice that this force is directly proportional to the velocity $v$ and acts in the opposite direction. This is the exact mathematical definition of a linear damping force, allowing us to use all the standard results from damped SHM theory.

Q: Does the damping affect the time period of the oscillation?

Strictly speaking, yes. The time period of a damped oscillator is slightly longer than its undamped counterpart, given by $T' = \frac{2\pi}{\sqrt{\omega_0^2 - (b/2m)^2}}$. However, in most practical problems like this one, the damping is "light" ($b/2m \ll \omega_0$), so we can safely approximate the time period using the undamped formula $T = 2\pi\sqrt{m/k}$.

Q: Why did we approximate $\pi$ as $\sqrt{10}$?

This is a very common and incredibly useful approximation in physics, especially in competitive exams like JEE. Since $\pi \approx 3.14159$, squaring it gives $\pi^2 \approx 9.8696$, which is very close to 10. Using this approximation often leads to clean integer cancellations, as it did in our time period calculation!

Q: Where does the kinetic energy of the strip go?

The mechanical energy (kinetic + elastic potential) of the oscillating strip is continuously converted into electrical energy due to the motional EMF. This electrical energy drives a current through the circuit's resistance, where it is ultimately dissipated as heat (Joule heating). This is why the amplitude decays over time!

Question

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 $0.5	ext{ Nm}^{-1}$ (see figure). The assembly is kept in a uniform magnetic field of $0.1	ext{ T}$. 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

Options:

(A) 1000

(B) 50000

(C) 5000

(D) 10000

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing right in front of this fascinating setup. We have a conducting strip, perfectly balanced on a U-shaped wire, connected to a spring. This entire assembly is bathed in a uniform magnetic field pointing into the screen.

When we pull the strip and release it, it doesn't just oscillate freely like a normal spring-mass system. As it moves with a velocity $v$, it cuts through the magnetic field lines. This motion induces a motional EMF across the strip, given by the famous equation $e = Blv$.

Because the U-shaped wire completes the circuit, this EMF drives an induced current $i = \frac{Blv}{R}$ through the loop. But nature loves a balance! According to Lenz's Law, this induced current will experience a magnetic Lorentz force $F_m = -Bil$ that opposes the very motion that created it.

### The Master Equation

Now, let's bring in Newton's second law to see how these forces dictate the strip's motion. The net force acting on the strip is the sum of the spring's restoring force and the opposing magnetic force.

$$m\frac{d^2x}{dt^2} = -kx - Bil$$

We know that the current $i$ is $\frac{Blv}{R}$, and the velocity $v$ is simply the rate of change of position, $\frac{dx}{dt}$. Let's substitute these into our equation:

$$m\frac{d^2x}{dt^2} = -kx - B\left(\frac{Bl}{R}\frac{dx}{dt}ight)l$$

Rearranging this, we get a beautiful, classic differential equation:

$$m\frac{d^2x}{dt^2} + \frac{B^2l^2}{R}\frac{dx}{dt} + kx = 0$$

I know this differential equation might look a bit terrifying at first glance, but let's take a breath. This is the exact mathematical signature of **damped simple harmonic motion**! The term $\frac{B^2l^2}{R}$ acts as our damping coefficient, $b$. The magnetic field is literally acting like a viscous fluid, draining energy from the system.

### Unraveling the Damping

In a damped harmonic oscillator, the amplitude doesn't stay constant; it decays exponentially over time. The formula for this decaying amplitude is:

$$A(t) = A_0 e^{-\frac{bt}{2m}}$$

The question asks for the number of oscillations before the amplitude decreases by a factor of $e$. This means we want to find the time $t$ when $A(t) = \frac{A_0}{e}$.

For this to happen, the exponent must be equal to 1:

$$\frac{bt}{2m} = 1 \Rightarrow t = \frac{2m}{b}$$

Substituting our damping coefficient $b = \frac{B^2l^2}{R}$, we get the expression for the total time:

$$t = \frac{2mR}{B^2l^2}$$

### Final Calculation

We are in the endgame now! Let's carefully plug in all the given values. We have $m = 50 	imes 10^{-3} 	ext{ kg}$, $R = 10\ \Omega$, $B = 0.1 	ext{ T}$, and $l = 0.1 	ext{ m}$.

$$t = \frac{2(50 	imes 10^{-3})(10)}{(0.1)^2(0.1)^2} = \frac{1000 	imes 10^{-3}}{10^{-4}} = 10000 	ext{ s}$$

So, it takes 10,000 seconds for the amplitude to drop by a factor of $e$. But the question asks for the **number of oscillations**, $N$. To find this, we need the time period of a single oscillation.

For light damping, the time period is practically identical to the undamped time period:

$$T = 2\pi\sqrt{\frac{m}{k}}$$

Substituting $m = 50 	imes 10^{-3} 	ext{ kg}$ and $k = 0.5 	ext{ N/m}$:

$$T = 2\pi\sqrt{\frac{50 	imes 10^{-3}}{0.5}} = 2\pi\sqrt{0.1} = \frac{2\pi}{\sqrt{10}}$$

Here is a pro-tip for JEE: $\pi^2$ is approximately 10, which means $\pi \approx \sqrt{10}$. Using this approximation, our time period simplifies beautifully to $T \approx 2 	ext{ s}$.

Finally, the number of oscillations is the total time divided by the time period:

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

And there we have it! The strip will perform **5000 oscillations** before its amplitude decays by a factor of $e$. A stunning interplay of mechanics and electromagnetism!

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The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation

Unraveling the Damping

Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced Current & Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Damped SHM Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Amplitude Decay

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time for Decay

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Oscillations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward