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 √3 2 𝑥̂ + 1 2 𝑦̂ 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 _______ m s −1 . [Given: The permeability of free space 𝜇0 = 4𝜋 × 10−7 N A −2 ]

To solve for the speed $v$ at which the rectangular conducting loop is moving away from the wire, we need to analyze the electromagnetic induction caused by the changing magnetic flux through the loop.

Step 1: Magnetic Field Due to the Wire

The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by:

$$B=\frac{{\mu }_{0}I}{2\pi r}$$

where:

• ${\mu }_0=4\pi \times 10^{-7}{\text{N/A}}^2$ is the permeability of free space,

• $I=10\text{A}$ is the current in the wire,

• $r$ is the distance from the wire.

Step 2: Magnetic Flux Through the Loop

The loop has a length $l=4\text{cm}=0.04\text{m}$ and a width $w=2\text{cm}=0.02\text{m}$. The loop is moving away from the wire in the direction $\frac{\sqrt{3}}{2}\widehat{x}+\frac{1}{2}\widehat{y}$, but the key is that the distance $r$ from the wire is changing.

The magnetic flux $\mathrm{\Phi }$ through the loop is:

$$\mathrm{\Phi }=B\cdot A$$

where $A=l\cdot w$ is the area of the loop. However, since the magnetic field $B$ varies with distance $r$, we need to consider the flux through a small strip of the loop at a distance $r$ from the wire.

The flux through a small strip of width $dr$ at a distance $r$ is:

$$d\mathrm{\Phi }=B\cdot l\cdot dr=\frac{{\mu }_{0}I}{2\pi r}\cdot l\cdot dr$$

Integrating from $r=d$ to $r=d+w$:

$$\mathrm{\Phi }={\int }_d^{d+w}\frac{{\mu }_{0}Il}{2\pi r}dr=\frac{{\mu }_{0}Il}{2\pi }\ln \left(\frac{d+w}{d}\right)$$

Step 3: Induced EMF and Current

The induced EMF $\mathcal{E}$ in the loop is given by Faraday's law:

$$\mathcal{E}=-\frac{d\mathrm{\Phi }}{dt}$$

The rate of change of flux is:

$$\frac{d\mathrm{\Phi }}{dt}=\frac{d}{dt}(\frac{{\mu }_{0}Il}{2\pi }\ln \left(\frac{d+w}{d}\right))$$

Since $d$ is changing with time as the loop moves away from the wire at speed $v$, we have:

$$\frac{d\mathrm{\Phi }}{dt}=\frac{{\mu }_{0}Il}{2\pi }\cdot \frac{d}{dt}(\ln \left(\frac{d+w}{d}\right))$$

Using the chain rule:

$$\frac{d\mathrm{\Phi }}{dt}=\frac{{\mu }_{0}Il}{2\pi }\cdot (\frac{1}{d+w}-\frac{1}{d})\cdot \frac{dd}{dt}$$

Since $\frac{dd}{dt}=v$, the speed at which the loop is moving away from the wire:

$$\frac{d\mathrm{\Phi }}{dt}=\frac{{\mu }_{0}Il}{2\pi }\cdot (\frac{1}{d+w}-\frac{1}{d})\cdot v$$

Simplifying:

$$\frac{d\mathrm{\Phi }}{dt}=\frac{{\mu }_{0}Il}{2\pi }\cdot \left(\frac{d-(d+w)}{d(d+w)}\right)\cdot v=\frac{{\mu }_{0}Il}{2\pi }\cdot \left(\frac{-w}{d(d+w)}\right)\cdot v$$

Thus, the induced EMF is:

$$\mathcal{E}=-\frac{d\mathrm{\Phi }}{dt}=\frac{{\mu }_{0}Ilw}{2\pi d(d+w)}\cdot v$$

Step 4: Relating EMF to Current

The induced current $I_{\text{loop}}$ in the loop is given by Ohm's law:

$$I_{\text{loop}}=\frac{\mathcal{E}}{R}$$

where $R=0.1\mathrm{\Omega }$ is the resistance of the loop. Given that $I_{\text{loop}}=10\mu \text{A}=10\times 10^{-6}\text{A}$, we have:

$$10\times 10^{-6}=\frac{\frac{{\mu }_{0}Ilw}{2\pi d(d+w)}\cdot v}{0.1}$$

Solving for $v$:

$$v=\frac{10\times 10^{-6}\times 0.1\times 2\pi d(d+w)}{{\mu }_{0}Ilw}$$

Substituting the given values:

$$v=\frac{10\times 10^{-6}\times 0.1\times 2\pi \times 0.04\times (0.04+0.02)}{4\pi \times 10^{-7}\times 10\times 0.04\times 0.02}$$

Simplifying:

$$v=\frac{10^{-6}\times 0.1\times 2\pi \times 0.04\times 0.06}{4\pi \times 10^{-7}\times 10\times 0.04\times 0.02}$$$$v=\frac{10^{-6}\times 0.1\times 0.06}{4\times 10^{-7}\times 10\times 0.02}$$$$v=\frac{0.006\times 10^{-6}}{8\times 10^{-7}}$$$$v=\frac{0.006}{0.8}=0.0075\text{m/s}$$

Thus, the speed $v$ is:

$$\boxed{{\displaystyle 0.0075\text{m/s}}}$$