A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v per

Q: Why doesn't the loop's movement create a current?

Even though the loop is moving, it is moving within a *uniform* magnetic field. This means the number of magnetic field lines passing through the loop remains exactly the same. Without a *change* in flux, there is no induced emf, and therefore no current.

Q: What if the loop was moving out of the magnetic field?

If the loop was exiting the magnetic field, the area of the loop exposed to the field would decrease. This would cause a decrease in magnetic flux, which *would* induce an emf and a current in the loop to oppose this change.

Q: Don't the moving charges in the wire experience a magnetic force?

Yes, they do! The charges in the leading and trailing vertical edges experience a magnetic force, which induces an emf ($e = BLv$) in each of those edges. However, because these two emfs are equal and oppose each other in the closed circuit, the net emf is zero.

Q: Would a current be induced if the loop was rotating instead of translating?

Absolutely. If the loop were rotating, the angle between the area vector and the magnetic field would change continuously. This would change the magnetic flux ($\Phi = BA \cos\theta$), thereby inducing an alternating current.

Q: Does the resistance $R$ of the loop matter in this specific case?

In this specific case, since the net induced emf is zero, the current is zero regardless of the resistance ($i = 0/R = 0$). The resistance would only matter for calculating the magnitude of the current if there were a non-zero net emf.

Question

A conducting square loop of side $L$ and resistance $R$ moves in its plane with a uniform velocity $v$ perpendicular to one of its sides. A magnetic induction $B$, constant in time and space, pointing perpendicular to and into the plane of the loop exists everywhere. The current induced in the loop is

Options:

(A) $BLv/R$ clockwise

(B) $BLv/R$ anti-clockwise

(C) $2BLv/R$ anti-clockwise

(D) zero

Sigma Percentile · Accepted Answer

The phenomenon of electromagnetic induction is one of the most beautiful and profound discoveries in physics. It connects electricity and magnetism in a dynamic dance. But sometimes, the most important lesson is understanding when this dance *doesn't* happen. Let's dive into this classic problem and uncover the physical reality behind it.

### Analyzing the Setup

Imagine you are observing a conducting square loop of side $L$. This loop is gliding smoothly with a constant velocity $v$ across a region of space. Now, this isn't just empty space; it's permeated by a magnetic field $B$.

The problem statement gives us a crucial piece of information: the magnetic field is **constant in time and space**. This means no matter where the loop goes within this region, and no matter how long you wait, the magnetic field it experiences remains exactly the same. It points uniformly into the plane of the loop.

### The Master Equation: Faraday's Law

To determine if a current is induced, we must consult the ultimate authority on the matter: **Faraday's Law of Electromagnetic Induction**. Faraday's law states that the induced electromotive force (emf) in a closed loop is directly proportional to the negative rate of change of magnetic flux through the loop.

Mathematically, this is expressed as:
$$e = -\frac{d\Phi}{dt}$$

Here, $\Phi$ represents the magnetic flux. The flux is essentially a count of how many magnetic field lines are piercing through the area of the loop. It is given by the dot product of the magnetic field vector and the area vector:
$$\Phi = \vec{B} \cdot \vec{A} = B A \cos(0^\circ) = B L^2$$

### The Crucial Evaluation

Now, let's look closely at our flux equation, $\Phi = B L^2$.

As the loop moves, does the magnetic field $B$ change? No, the problem explicitly states it is constant in space. 
Does the area $A = L^2$ change? No, the loop is a rigid square; it's not expanding or shrinking.

Since both $B$ and $A$ are perfectly constant as the loop translates, the total magnetic flux $\Phi$ passing through the loop is also a constant value.

What happens when we take the time derivative of a constant? 
$$\frac{d\Phi}{dt} = \frac{d}{dt}(B L^2) = 0$$

Because the rate of change of flux is zero, Faraday's law tells us that the induced emf $e$ must be exactly zero.

### The Motional EMF Perspective

We can also arrive at this conclusion by looking at the microscopic forces acting on the charges within the wire, a concept known as **motional emf**.

As the loop moves to the right, the vertical wires (the leading and trailing edges) are cutting through the magnetic field lines. According to the Lorentz force law, the free electrons in these vertical segments experience a force, which induces an emf in each segment.

For a straight wire of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$, the induced motional emf is:
$$e = BLv$$

So, the leading edge acts like a battery with emf $e_1 = BLv$. The trailing edge, moving with the exact same velocity through the exact same field, also acts like a battery with emf $e_2 = BLv$.

However, because these two "batteries" are connected in a closed loop and are pushing charge in opposite directions around the loop, they perfectly oppose each other. 
$$e_{	ext{net}} = e_1 - e_2 = BLv - BLv = 0$$

### Final Calculation

Whether we use Faraday's law of flux or the concept of motional emf, the conclusion is inescapable. The net electromotive force driving the circuit is zero.

By Ohm's law, the induced current $i$ is the net emf divided by the resistance $R$:
$$i = \frac{e_{	ext{net}}}{R} = \frac{0}{R} = 0$$

The loop glides through the uniform magnetic field silently, with absolutely zero current induced within it. This problem serves as a powerful reminder: it is not the *presence* of a magnetic field or even the *motion* through it that induces a current, but strictly the **change** in the magnetic flux!

Select Answer:

Visualized Solution

\text{Analyzing the Setup}

\text{Magnetic Flux}

\text{Faraday's Law of Induction}

\text{Rate of Change of Flux}

\text{Motional EMF Perspective}

\text{Final Induced Current}

\text{The Way Forward}

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

The Crucial Evaluation

The Motional EMF Perspective

Final Calculation

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