A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity v0 in

Q: Why doesn't the right arm of the loop generate an EMF?

The right arm is outside the magnetic field region. Motional EMF is only generated when a conductor actively cuts through magnetic field lines.

Q: How do we know the Wheatstone bridge is balanced?

The ratio of the resistances in the adjacent arms is equal ($3/3 = 3/3$). Therefore, the bridge is balanced and the central resistor carries no current.

Q: Why do we add the loop's resistance to the network's equivalent resistance?

The loop itself acts as the voltage source (battery) for the circuit, and its resistance acts as the internal resistance of this battery. They are in series.

Q: What if the loop was moving entirely inside the magnetic field?

If the entire loop was inside, both the left and right arms would generate equal and opposite EMFs, cancelling each other out. The net EMF and current would be zero.

Q: How exactly does Lenz's law give us a clockwise direction?

As the loop exits the field, the number of magnetic field lines pointing *into* the page decreases. To oppose this loss, the induced current creates its own field pointing *into* the page. Using the right-hand rule, your thumb points into the page, and your fingers curl clockwise.

Question

A square metal wire loop of side 10 cm and resistance 1 $\Omega$ is moved with a constant velocity $v_0$ in a uniform magnetic field of induction $B = 2 	ext{ Wb/m}^2$ as shown in the figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 $\Omega$.
What should be the speed of the loop so as to have a steady current of 1mA in the loop? Give the direction of current in the loop.

Sigma Percentile · Accepted Answer

## Motional EMF Meets the Wheatstone Bridge: A Tale of Moving Loops

Imagine you are pulling a metal frame out of a dense magnetic field. As you pull, you are not just moving metal; you are forcing electrons to march, creating a current out of thin air. This is the magic of **motional EMF**. In this problem, we combine this beautiful phenomenon with a classic circuit puzzle: the Wheatstone bridge.

### Analyzing the Setup

Let's look at the physical reality of our system. We have a square loop of side $10	ext{ cm}$ (or $0.1	ext{ m}$) moving to the right with a constant velocity $v_0$. The magnetic field $B$ is uniform and points directly into the page, but it only exists on the left side of the dashed boundary.

As the loop moves, its right arm is already outside the magnetic field. It cuts no flux lines, so it generates no voltage. However, the **left arm** is still inside the field. As it moves to the right, it slices through the magnetic field lines. This cutting action generates a motional EMF, effectively turning the left arm into a battery with voltage $e = B v_0 l$.

### The Resistor Network

Now, let's shift our focus to the network of resistors connected to the loop. At first glance, it might look like a complex web, but if you trace the connections, a familiar shape emerges: a **Wheatstone bridge**.

The bridge consists of four outer resistors, each $3\,\Omega$, and one central resistor, also $3\,\Omega$. Because the ratio of the adjacent arms is equal ($\frac{3}{3} = \frac{3}{3}$), the bridge is perfectly balanced.

What does a balanced bridge mean for us? It means the potential difference across the central resistor is zero. No current flows through it, so we can completely ignore it! The equivalent resistance of the remaining network is simply two parallel branches of $6\,\Omega$ each, which gives us:

$$R_{eq} = \frac{6 	imes 6}{6 + 6} = 3\,\Omega$$

Don't forget the loop itself! The loop has a resistance of $1\,\Omega$, which acts like the internal resistance of our "motional battery." Therefore, the total resistance of the entire circuit is:

$$R_{total} = 1\,\Omega + 3\,\Omega = 4\,\Omega$$

### The Master Equation

We are given that the steady current $i$ in the loop must be $1	ext{ mA}$, or $10^{-3}	ext{ A}$. According to Ohm's law, the current is the total EMF divided by the total resistance:

$$i = \frac{e}{R_{total}} = \frac{B v_0 l}{R_{total}}$$

Let's substitute the values we know into this master equation:

$$10^{-3} = \frac{2 \cdot v_0 \cdot 0.1}{4}$$

### Final Calculation

Now, it's just a matter of simple algebra. Let's isolate $v_0$:

$$10^{-3} = \frac{0.2 \cdot v_0}{4}$$

$$10^{-3} = 0.05 \cdot v_0$$

$$v_0 = \frac{10^{-3}}{0.05} = 0.02\,	ext{m/s}$$

So, you need to pull the loop at exactly $0.02	ext{ m/s}$ to maintain that $1	ext{ mA}$ current.

### Finding the Direction

Finally, which way does the current flow? We turn to **Lenz's Law**, which states that nature abhors a change in flux.

As the loop moves out of the magnetic field, the number of magnetic field lines pointing *into* the page is decreasing. To fight this loss, the induced current will try to create its own magnetic field pointing *into* the page.

Using the right-hand grip rule, point your right thumb into the screen. Your fingers will naturally curl in a **clockwise** direction. Thus, the induced current flows clockwise around the loop.

Visualized Solution

$Visualizing the Setup$

$Motional EMF$

$Wheatstone Bridge$

$Equivalent Resistance$

$Ohm’s Law$

$Substitution$

$Calculating Speed$

$Direction of Current$

$The Way Forward$

The Sigma Insight: Motional EMF

Analyzing the Setup

The Resistor Network

The Master Equation

Final Calculation

Finding the Direction

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A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R Ω$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_{1}, I_{2}$ and $I$ are

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm’s Law

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

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

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

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