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 3 2 x ∧ + 1 2

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 $\frac{\sqrt{3}}{2} \overset{\land}{x} + \frac{1}{2} \overset{\land}{y}$ 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 _____ $m s^{- 1}$ .

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

TestHub · Accepted Answer

The two sides of the rectangular loop which are perpendicular to the wire would not contribute in induced emf.
Angle of velocity vector with x-axis is $\tan α = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$ . Therefore, $α = 30 °$ .
For parallel sides:
$\vec{E} = \vec{B} \times \vec{v}$
$= \frac{μ_{0} I}{2 π x} \times v$
$\Rightarrow$ Net emf $= (E_{1} \cos 60^{o} - E_{2} \cos 60^{o}) \times$ width
$= \frac{1}{2} \times \frac{2}{100} \times \frac{μ_{0} I v}{2 π} [\frac{1}{\frac{4}{100}} - \frac{1}{\frac{8}{100}}]$
$= \frac{1}{100} \times (2 \times 10^{- 7}) \times 10 \times v \times \frac{100}{8}$
$= 2.5 v \times 10^{- 7}$
Now, net EMF $= i \times R$
$\Rightarrow v = \frac{(10 \times 10^{- 6}) \times 0.1}{2.5 \times 10^{- 7}} = 4 m s^{- 1}$

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