A rectangular conducting loop of length 4 cm and width 2 cm is in the xy-plane, as shown in the figu

Q: Why doesn't the horizontal velocity $v_x$ induce a net current?

While $v_x$ does induce an EMF in the vertical sides of the loop, both vertical sides are at the exact same distance from the wire at any given moment. Therefore, they experience the same magnetic field and generate equal but opposing EMFs that perfectly cancel each other out.

Q: Why do we ignore the EMF induced in the vertical sides by the vertical velocity $v_y$?

The motional EMF is calculated using $(\vec{v} \times \vec{B}) \cdot d\vec{l}$. For the vertical velocity $v_y$, the cross product $\vec{v}_y \times \vec{B}$ points horizontally. Since the vertical sides are oriented vertically, the dot product with $d\vec{l}$ is zero. No EMF is induced along the vertical sides by vertical motion.

Q: Do I need to convert the centimeters to meters before calculating?

Technically yes, for standard SI unit consistency. However, if you look closely at the formula, the width $b$ (in the numerator) and the distances $d$ and $d+a$ (in the denominator) all have units of length. The $10^{-2}$ conversion factors perfectly cancel out, leaving a dimensionless ratio.

Q: How did we get the angle for the velocity?

The direction vector is given as $\frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y}$. This is a unit vector where the y-component is $\sin(\theta) = 1/2$. This corresponds to an angle of $30^\circ$ with the x-axis.

Q: What if the loop was moving towards the wire instead of away?

If the loop was moving towards the wire, the magnetic flux through the loop would be increasing rather than decreasing. According to Lenz's Law, the induced current would reverse its direction to oppose this change.

Question

A rectangular conducting loop of length $4	ext{ cm}$ and width $2	ext{ cm}$ is in the xy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction $\frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y}$ with a constant speed $v$. The wire is carrying a steady current $I = 10	ext{ A}$ in the positive x-direction. A current of $10	ext{ }\mu	ext{A}$ flows through the loop when it is at a distance $d = 4	ext{ cm}$ from the wire. If the resistance of the loop is $0.1	ext{ }\Omega$, then the value of $v$ is _________ $	ext{ms}^{-1}$.

[Given: The permeability of free space $\mu_0 = 4\pi 	imes 10^{-7}	ext{ NA}^{-2}$]

Sigma Percentile · Accepted Answer

### The Setup
Imagine a rectangular conducting loop placed near a long, straight wire carrying a steady current $I = 10	ext{ A}$. The loop is moving away from the wire at a constant speed $v$, but not straight away—it's moving at an angle. The velocity vector is given as $\vec{v} = v\left(\frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y}ight)$.

This tells us that the loop has both a horizontal velocity component $v_x = v\frac{\sqrt{3}}{2}$ and a vertical velocity component $v_y = v\frac{1}{2}$. The angle of motion is $30^\circ$ with respect to the x-axis.

### The Magnetic Field
Before we calculate any induced EMF, we must understand the environment the loop is moving through. The straight wire creates a magnetic field that points out of the page (in the $+\hat{k}$ direction) in the region where the loop is located.

The strength of this magnetic field is not uniform; it decreases as you move further away from the wire according to Ampere's Law: 
$$B = \frac{\mu_0 I}{2\pi y}$$

### Deconstructing Velocity
When dealing with motional EMF, the formula is $\varepsilon = \int (\vec{v} 	imes \vec{B}) \cdot d\vec{l}$. Let's analyze the effect of the two velocity components separately.

**The Horizontal Velocity ($v_x$):**
As the loop moves horizontally, the two vertical sides cut through the magnetic field lines. This induces an EMF in both vertical sides. However, because both vertical sides are at the exact same vertical distance $y$ from the wire at any given instant, they experience the exact same magnetic field strength. Consequently, the EMF induced in the left side perfectly cancels the EMF induced in the right side. The horizontal velocity $v_x$ contributes **zero** to the net EMF of the loop!

**The Vertical Velocity ($v_y$):**
As the loop moves vertically, the horizontal sides cut through the magnetic field lines. The bottom side of the loop is at a distance $d$ from the wire, while the top side is at a distance $d+a$. Because the bottom side is closer to the wire, it experiences a stronger magnetic field ($B_{	ext{bottom}}$) than the top side ($B_{	ext{top}}$). This difference in field strength creates a net EMF that drives the current around the loop.

### The Master Equation
The EMF induced in the bottom side is $\varepsilon_1 = B_{	ext{bottom}} b v_y$, and the EMF induced in the top side is $\varepsilon_2 = B_{	ext{top}} b v_y$. Since they oppose each other, the net EMF is:
$$\varepsilon_{	ext{net}} = \varepsilon_1 - \varepsilon_2 = (B_{	ext{bottom}} - B_{	ext{top}}) b v_y$$

Substituting the magnetic field formula, we get:
$$\varepsilon_{	ext{net}} = \frac{\mu_0 I}{2\pi} \left( \frac{1}{d} - \frac{1}{d+a} ight) b v_y$$

By Ohm's Law, the induced current is $i = \frac{\varepsilon_{	ext{net}}}{R}$.

### Final Calculation
Now, we carefully substitute the given values: $i = 10	ext{ }\mu	ext{A} = 10^{-5}	ext{ A}$, $R = 0.1	ext{ }\Omega$, $b = 2	ext{ cm} = 2 	imes 10^{-2}	ext{ m}$, $d = 4	ext{ cm} = 4 	imes 10^{-2}	ext{ m}$, and $a = 4	ext{ cm} = 4 	imes 10^{-2}	ext{ m}$.

$$10^{-5} = \frac{4\pi 	imes 10^{-7} 	imes 10}{2\pi 	imes 0.1} \left( \frac{1}{4 	imes 10^{-2}} - \frac{1}{8 	imes 10^{-2}} ight) 	imes (2 	imes 10^{-2}) 	imes v_y$$

Notice how the $10^{-2}$ from the width $b$ perfectly cancels the $10^{-2}$ from the distances in the denominator! This leaves us with a clean, dimensionless fraction:
$$10^{-5} = \frac{2 	imes 10^{-6} 	imes 10}{0.1} \left( \frac{1}{4} - \frac{1}{8} ight) 	imes 2 	imes v_y$$
$$10^{-5} = 2 	imes 10^{-4} 	imes \frac{1}{8} 	imes 2 	imes v_y$$
$$10^{-5} = 4 	imes 10^{-5} 	imes \frac{1}{8} 	imes v_y$$
$$10^{-5} = \frac{1}{2} 	imes 10^{-5} 	imes v_y$$

Solving this yields $v_y = 2	ext{ m/s}$.

Finally, we relate the vertical velocity back to the total speed $v$. From our initial velocity vector, we know that $v_y = v \frac{1}{2}$.
$$2 = \frac{v}{2} \implies v = 4	ext{ m/s}$$

The loop is moving at a constant speed of **$4	ext{ m/s}$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$The Magnetic Field$

$Deconstructing Velocity$

$EMF on Vertical Sides$

$EMF on Horizontal Sides$

$The Master Equation$

$Substituting Values$

$Solving for v_{y}$

$Final Speed$

$The Way Forward$

The Sigma Insight: Motional EMF

The Setup Imagine a rectangular conducting loop placed near a long, straight wire carrying a steady current $I = 10 A$

The Magnetic Field Before we calculate any induced EMF, we must understand the environment the loop is moving through

Deconstructing Velocity When dealing with motional EMF, the formula is $ε = \int (v \times B) \cdot d l$

The Master Equation The EMF induced in the bottom side is $ε_{1} = B_{bottom} b v_{y}$ , and the EMF induced in the top side is $ε_{2} = B_{top} b v_{y}$

Final Calculation

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Enter Numerical Value:

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; } The Magnetic Field

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

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

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

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

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

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

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

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