A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges

Q: Why is the magnetic force only acting on the leading edge?

While the loop is entering, only the leading edge is cutting the magnetic field lines. The side edges are parallel to the velocity, so they don't generate motional EMF, and the trailing edge is still outside the magnetic field.

Q: Why does the force become zero when the loop is completely inside?

Once the loop is fully inside the uniform magnetic field, the magnetic flux through it becomes constant. According to Faraday's Law, a constant flux means zero induced EMF, which means zero current and therefore zero magnetic force.

Q: Why doesn't the loop ever come to a complete stop in finite time?

The magnetic force is proportional to the velocity ($F \propto v$). As the loop slows down, the opposing force also decreases. This leads to an exponential decay in velocity ($v = v_0 e^{-Kt}$), meaning it mathematically approaches zero only as time approaches infinity.

Q: How did we get $a = -Kv$?

By equating Newton's second law ($F = Ma$) with the magnetic force ($F = - \frac{B_0^2 L^2}{R} v$). Dividing by mass $M$ gives $a = - \left(\frac{B_0^2 L^2}{RM}\right) v$. The term in the parentheses is defined as $K$ in the problem.

Q: Can we use energy conservation here?

Yes! The loss in kinetic energy of the loop is exactly equal to the heat dissipated in the resistor due to the induced current. However, to find the exact time or position, solving the differential equation of motion is much more direct.

Question

A conducting square loop of side $L$, mass $M$ and resistance $R$ is moving in the $XY$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \ge 0$ has a uniform magnetic field, $\vec{B} = B_0 \hat{k}$. The magnetic field is zero everywhere else. At time $t = 0$, the loop starts to enter the magnetic field with an initial velocity $v_0 \hat{j}$ m/s, as shown in the figure. Considering the quantity $K = \frac{B_0^2 L^2}{RM}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

Options:

(A) If $v_0 = 1.5KL$, the loop will stop before it enters completely inside the region of magnetic field.

(B) When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

(C) If $v_0 = \frac{KL}{10}$, the loop comes to rest at $t = \left(\frac{1}{K}\right) \ln \left(\frac{5}{2}\right)$.

(D) If $v_0 = 3KL$, the complete loop enters inside the region of magnetic field at time $t = \left(\frac{1}{K}\right) \ln \left(\frac{3}{2}\right)$.

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Loop Entering the Field

Imagine a conducting square loop moving steadily towards a region filled with a uniform magnetic field. As soon as the leading edge of the loop crosses the boundary into the field, things start to get interesting. The magnetic flux passing through the loop begins to change. According to Faraday's Law of Induction, this changing flux induces an electromotive force (EMF) in the loop.

### The Physics: Motional EMF and Magnetic Force

Only the leading edge of the loop is actively cutting the magnetic field lines. The motional EMF generated is given by the classic formula:

$$ \varepsilon = B_0 L v $$

This EMF acts like a battery, driving a current $i$ through the loop's resistance $R$:

$$ i = \frac{B_0 L v}{R} $$

Now, we have a current-carrying wire (the leading edge) sitting inside a magnetic field. The Lorentz force kicks in! The magnetic force $F_m$ acting on this edge is:

$$ F_m = i L B_0 = \frac{B_0^2 L^2 v}{R} $$

By Lenz's Law, this force must oppose the change that created it. Therefore, the magnetic force acts in the opposite direction of the velocity, acting as a magnetic brake.

### The Math: Differential Equations of Motion

Let's apply Newton's Second Law. The only horizontal force acting on the loop is this magnetic braking force. So, we can write:

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

The problem conveniently defines a constant $K = \frac{B_0^2 L^2}{RM}$. Substituting this in, our acceleration simplifies beautifully to:

$$ a = -K v $$

This is a fundamental differential equation. We can analyze it in two ways: with respect to position ($y$) and with respect to time ($t$).

**1. Velocity as a function of position:**
We use the chain rule trick $a = v \frac{dv}{dy}$:

$$ v \frac{dv}{dy} = -K v \implies \frac{dv}{dy} = -K $$

Integrating this from $y=0$ to $y$ gives a linear relationship:

$$ v(y) = v_0 - K y $$

**2. Velocity and Position as a function of time:**
We use the standard definition $a = \frac{dv}{dt}$:

$$ \frac{dv}{dt} = -K v \implies \int \frac{dv}{v} = -K \int dt $$

This yields an exponential decay for velocity:

$$ v(t) = v_0 e^{-Kt} $$

Integrating velocity gives us the position over time:

$$ y(t) = \frac{v_0}{K} (1 - e^{-Kt}) $$

### The Analysis: Testing the Options

Now we are fully equipped to test the given options.

**Option A:** The loop enters completely when its trailing edge crosses the boundary, which means the leading edge is at $y = L$. If $v_0 = 1.5KL$, the velocity at $y = L$ is $v(L) = 1.5KL - K(L) = 0.5KL$. Since the velocity is still positive, the loop does not stop before entering completely. Option A is incorrect.

**Option B:** Once the loop is completely inside the uniform magnetic field, the magnetic flux through it becomes constant. A constant flux means zero induced EMF, zero current, and consequently, zero magnetic force. Option B is correct.

**Option C:** The velocity function $v(t) = v_0 e^{-Kt}$ is an exponential decay. Mathematically, it never reaches exactly zero in finite time. Therefore, the loop never truly "comes to rest" at a specific finite time. Option C is incorrect.

**Option D:** We want to find the time $t$ when the loop enters completely ($y = L$) given $v_0 = 3KL$. We plug these into our position-time equation:

$$ L = \frac{3KL}{K} (1 - e^{-Kt}) $$

$$ L = 3L (1 - e^{-Kt}) $$

$$ \frac{1}{3} = 1 - e^{-Kt} \implies e^{-Kt} = \frac{2}{3} \implies e^{Kt} = \frac{3}{2} $$

Taking the natural logarithm of both sides gives:

$$ t = \frac{1}{K} \ln\left(\frac{3}{2}\right) $$

This matches the option perfectly! Option D is correct.

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Motional EMF}

\text{Magnetic Force}

\text{Equation of Motion}

\text{Acceleration}

\text{Velocity vs Position}

\text{Evaluating Options A \& B}

\text{Velocity vs Time}

\text{Position vs Time}

\text{Evaluating Options C \& D}

The Sigma Insight: Motional EMF

The Setup

The Physics

The Math

The Analysis

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