A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axi

Q: Why does the velocity decrease linearly with position $x$ instead of time $t$?

The magnetic force is proportional to velocity ($F \propto -v$). Using the chain rule for acceleration, $a = v \frac{dv}{dx}$, we get $v \frac{dv}{dx} \propto -v$. Canceling $v$ from both sides leaves $\frac{dv}{dx} = \text{constant}$, which means velocity decreases linearly with position $x$.

Q: Why is the current zero when the loop is fully inside the magnetic field?

When the loop is completely inside the uniform magnetic field, the total magnetic flux passing through it remains constant as it moves. Since induced EMF depends on the rate of change of flux, a constant flux means zero EMF and therefore zero current.

Q: Why does the current change sign when exiting the field?

When entering, the inward magnetic flux is increasing, so Lenz's Law dictates an induced current that creates an outward field (counter-clockwise, positive). When exiting, the inward flux is decreasing, so the induced current creates an inward field (clockwise, negative) to oppose the decrease.

Q: Why is the force always negative (opposing motion) in both entering and exiting phases?

This is a direct consequence of the conservation of energy and Lenz's Law. The induced current will always flow in a direction such that the resulting magnetic force opposes the cause of the induction—in this case, the forward motion of the loop.

Q: What does 'sufficiently large $v_0$' mean in the problem statement?

It ensures that the loop has enough initial kinetic energy to completely pass through the magnetic field region without coming to a halt inside it due to the magnetic damping force.

Question

A rigid wire loop of square shape having side of length $L$ and resistance $R$ is moving along the $x$-axis with a constant velocity $v_0$ in the plane of the paper. At $t = 0$, the right edge of the loop enters a region of length $3L$, where there is a uniform magnetic field $B_0$ into the plane of the paper, as shown in the figure. For sufficiently large $v_0$, the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v(x)$, $I(x)$ and $F(x)$ represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of $x$. Counter-clockwise current is taken as positive.

Which of the following schematic plot(s) is (are) correct? (Ignore gravity)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a square loop of wire, armed with a resistance $R$, sliding smoothly along the $x$-axis. It's moving with an initial constant velocity $v_0$ right before it hits a region filled with a uniform magnetic field $B_0$ pointing into the page.

The magnetic field stretches from $x = 0$ to $x = 3L$. Our mission is to track the loop's velocity $v(x)$ and the induced current $I(x)$ as its right edge, located at $x$, travels through and eventually exits this magnetic zone.

To make sense of this, we will break the loop's journey into three distinct phases: entering the field, moving fully inside the field, and finally, exiting the field.

### Phase 1: Entering the Magnetic Field

As the right edge of the loop crosses $x = 0$ and moves up to $x = L$, the loop is partially inside the magnetic field.

The area of the loop exposed to the magnetic field is $L \cdot x$. Therefore, the magnetic flux $\phi$ pointing into the page is increasing:

$$ \phi = B_0 L x $$

According to **Faraday's Law** and **Lenz's Law**, nature hates a change in flux. To oppose this increasing inward flux, the loop induces an electromotive force (EMF) that drives a counter-clockwise current. Since the problem defines counter-clockwise as positive, our current $I$ is positive:

$$ I = \frac{1}{R} \frac{d\phi}{dt} = \frac{B_0 L v}{R} $$

This induced current interacts with the external magnetic field, creating a magnetic force $F$ that opposes the loop's motion:

$$ F = -I L B_0 = -\frac{B_0^2 L^2 v}{R} $$

Using Newton's second law, we can relate this force to the loop's acceleration. By applying the chain rule, $a = v \frac{dv}{dx}$, we get:

$$ m v \frac{dv}{dx} = -\frac{B_0^2 L^2 v}{R} $$

Notice how the velocity $v$ beautifully cancels out on both sides! This leaves us with a constant spatial deceleration:

$$ \frac{dv}{dx} = -\frac{B_0^2 L^2}{mR} $$

This tells us that **the velocity $v(x)$ decreases linearly** with position $x$. And since the current $I(x)$ is directly proportional to $v(x)$, **the current $I(x)$ also decreases linearly** from its initial positive value.

### Phase 2: Fully Inside the Field

Once the right edge passes $x = L$, the entire loop is submerged in the uniform magnetic field. This phase lasts until the right edge reaches $x = 3L$.

During this time, the total magnetic flux passing through the loop is constant:

$$ \phi = B_0 L^2 = 	ext{constant} $$

Because the flux isn't changing, the rate of change of flux is zero. Consequently, there is no induced EMF and **the induced current drops to zero**:

$$ I = 0 $$

With no current, there is no magnetic opposing force ($F = 0$). Therefore, the loop coasts through this region, and **its velocity $v(x)$ remains perfectly constant**.

### Phase 3: Exiting the Field

The final act begins when the right edge crosses $x = 3L$. The loop starts to exit the magnetic field, and this phase continues until $x = 4L$.

Now, the area of the loop still inside the field is shrinking. The inward magnetic flux is decreasing:

$$ \phi = B_0 L (4L - x) $$

To oppose this decrease, Lenz's Law dictates that the induced current must create its own inward magnetic field. This requires a clockwise current. Since counter-clockwise is positive, our current is now negative:

$$ I = -\frac{B_0 L v}{R} $$

Once again, this current produces a magnetic force that opposes the forward motion of the loop:

$$ F = -\frac{B_0^2 L^2 v}{R} $$

Following the exact same kinematic logic as in Phase 1, we find the spatial deceleration:

$$ \frac{dv}{dx} = -\frac{B_0^2 L^2}{mR} $$

So, **the velocity $v(x)$ continues to decrease linearly** with the exact same slope as before.

What about the current? The current is negative and proportional to velocity ($I \propto -v$). Since $v$ is decreasing linearly, the magnitude of the current decreases linearly. This means **the current $I(x)$ starts at a negative value and rises linearly towards zero**.

### The Final Verdict

By piecing together the physics of these three phases, we can perfectly visualize the graphs.

The velocity $v(x)$ drops linearly, stays flat, and then drops linearly again. The current $I(x)$ starts positive and drops linearly, flatlines at zero, and then jumps to a negative value before rising linearly back to zero.

Comparing our derived behavior with the given options, we can confidently conclude that the schematic plots for current and velocity match perfectly with **options (b) and (c)**.

Select Answer:

Visualized Solution

$x \in [0, 4 L]$

$0 < x < L$

$F = - I L B_{0}$

$L < x < 3 L$

$3 L < x < 4 L$

$F = - \frac{B _{0}^{2} L ^{2} v}{R}$

\text{Final Answer}

The Sigma Insight: Motional EMF

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Verdict

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