A rectangular conducting loop of length 4 cm and width 2 cm is in the -plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction with a constant speed . The wire is carrying a steady current in the po

Question

A rectangular conducting loop of length 4 cm and width 2 cm is in the $x y$ -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 $I = 10 A$ in the positive $x$ -direction. A current of $10 μ A$ 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 ______ ${ms}^{- 1}$ .

[Given: The permeability of free space $μ_{0} = 4 π \times 10^{- 7} {N A}^{- 2}$ ] [2023]

Smart Achievers · Accepted Answer

(4)

$\frac{\sqrt{3}}{2} \hat{x} + \frac{1}{2} \hat{y}$

$Current, i = \frac{ε}{R} = \frac{μ_{0} I}{2 π R} (\frac{1}{d} - \frac{1}{d + a}) b v_{y}$

$μ_{0} = 4 π \times 10^{- 7} {N A}^{- 2}$

$[Given: I = 10 A, R = 0.1 Ω, i = 10 μ A = 10^{- 5} A]$

$∴$ $v_{y} = 2$

$\tan θ = \frac{v_{y}}{v_{x}} \Rightarrow \frac{1}{\sqrt{3}} = \frac{2}{v_{x}}$

$∴$ $v_{x} = 2 \sqrt{3}$

$∴$ $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(2 \sqrt{3})}^{2} + {(2)}^{2}}$

$= \sqrt{16} = 4 {ms}^{- 1}$

Solution

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Solution

Ans. (4) Here e.m.f., ε=(B1-B2)bvy Current, i=εR=μ0I2πR(1d-1d+a)bvy ⇒10-5=2×10-7×100.1[14-18]×2·vy [Given: I=10 A, R=0.1 Ω, i=10 μA=10-5 A] ∴ vy=2 tanθ=vyvx⇒13=2vx ∴ vx=23 ∴ v=vx2+vy2=(23)2+(2)2 =16=4 ms-1

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