A rectangular conducting loop of length 4 cm and width 2 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 $\frac{\sqrt{3}}{2}\hat{x}+\frac{1}{2}$ with a constant speed v. The wire is carrying a steady current I = 10 A in the positive x-direction. A current of 10 µA flows through the loop when it is at a distance d = 4 cm from the wire. If the resistance of the loop is 0.1 Ω, then the value of v is .

[Given: The permeability of free space ${\mu }_0=4\pi \times {10}^{-7}{\text{ N A}}^{-2}$]

The correct answer is 4 m/s (or 4).

Step 1: Understand the given setup from the problem and diagram
From the given text and image, we have a rectangular loop moving in the xy-plane near a long straight conducting wire carrying a current $I=10\text{ A}$ along the positive x-direction.
• Length of the loop along the y-axis, $L=4\text{ cm}=0.04\text{ m}$
• Width of the loop along the x-axis, $w=2\text{ cm}=0.02\text{ m}$
• Distance of the bottom edge of the loop from the wire, $d=4\text{ cm}=0.04\text{ m}$
• Resistance of the loop, $R=0.1\text{ Ω}$
• Induced current in the loop, $i=10\text{ µA}={10}^{-5}\text{ A}$
• Velocity of the loop, $\overrightarrow{v}=v\left(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\right)$

Step 2: Determine the induced electromotive force (emf)
The induced emf $E$ in the loop is related to the induced current $i$ and resistance $R$ by Ohm's Law:

$E=i\times R=\left({10}^{-5}\text{ A}\right)\times \left(0.1\text{ Ω}\right)={10}^{-6}\text{ V}$

Step 3: Calculate the motional emf across each side of the loop
The magnetic field due to an infinitely long current-carrying wire at a distance $y$ from the wire is directed perpendicular to the xy-plane (into the page, along $-\hat{k}$):

$\overrightarrow{B}\left(y\right)=-\frac{{\mu }_{0}I}{2\pi y}\hat{k}$

The motional emf induced in a segment of length $\overrightarrow{dl}$ moving with velocity $\overrightarrow{v}$ is given by $E=\int \left(\overrightarrow{v}\times \overrightarrow{B}\right)·\overrightarrow{dl}$.
Since $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$, we have:

$\overrightarrow{v}\times \overrightarrow{B}=\left(v_x\hat{i}+v_y\hat{j}\right)\times \left(-B\hat{k}\right)=v_{x}B\hat{j}-v_{y}B\hat{i}$

• For the two vertical sides (parallel to the y-axis), motion along the x-direction ($v_x$) produces no net emf because the magnetic field does not vary with x, so the contributions from both vertical sides cancel out completely.
• For the horizontal sides (parallel to the x-axis), the integration along $\hat{i}$ picks up the component $-v_{y}B$.
Therefore, only the vertical component of velocity $v_y=\frac{v}{2}$ contributes to the net induced emf across the bottom and top edges of the loop:

$E=\left(B_{\text{bottom}}-B_{\text{top}}\right)·w·v_y$

Step 4: Express magnetic fields at the bottom and top edges
• Distance of bottom edge from wire, $y_1=d=4\text{ cm}=0.04\text{ m}$
• Distance of top edge from wire, $y_2=d+L=4\text{ cm}+4\text{ cm}=8\text{ cm}=0.08\text{ m}$

$B_{\text{bottom}}-B_{\text{top}}=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{y_1}-\frac{1}{y_2}\right)$

Substitute the given numerical values:

$\frac{{\mu }_{0}I}{2\pi }=\frac{\left(4\pi \times {10}^{-7}\right)\times 10}{2\pi }=2\times {10}^{-6}\text{ T m}$

$\left(\frac{1}{0.04}-\frac{1}{0.08}\right)=25-12.5=12.5{\text{m}}^{-1}$

$B_{\text{bottom}}-B_{\text{top}}=\left(2\times {10}^{-6}\right)\times 12.5=2.5\times {10}^{-5}\text{ T}$

Step 5: Solve for the velocity v
Substitute all calculated values into the emf equation:

$E=\left(B_{\text{bottom}}-B_{\text{top}}\right)·w·\left(\frac{v}{2}\right)$

${10}^{-6}=\left(2.5\times {10}^{-5}\right)\times 0.02\times \frac{v}{2}$

${10}^{-6}=\left(5\times {10}^{-7}\right)\times \frac{v}{2}$

${10}^{-6}=2.5\times {10}^{-7}\times v$

$v=\frac{{10}^{-6}}{2.5\times {10}^{-7}}=4\text{ m/s}$

Thus, the value of $v$ is 4.