Question

A rectangular conducting loop of length 4 cm and width 2 cm is in the X-Y plane. It is being moved away from a thin and long conducting wire along the direction (\(\frac{\sqrt3}{2}\hat{x}+\frac{1}{2}\hat{y}\)) with constant speed v. The wire is carrying a steady current I = 10 A in the +ve 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 \(\Omega\). Then the value of v is ........m/s

Answer 1

Correct Answer: 4

EMF Calculation Due to a Moving Conductor Near a Wire 

The two sides perpendicular to the wire would contribute net zero emf.

For parallel sides:

The electric field induced is given by: \[ \vec{E} = \vec{B} \times \vec{v} \] Using Biot-Savart Law: \[ \vec{E} = \frac{\mu_0 I}{2 \pi x} \times v \]

Therefore, the net EMF between the two sides is: \[ \text{Net emf} = \left(E_1 \cos 60^\circ - E_2 \cos 60^\circ\right) \times \text{width} \] Substituting values: \[ = \frac{1}{2} \times \frac{2}{100} \times \frac{\mu_0 I v}{2\pi} \left[\frac{1}{4} - \frac{1}{8}\right] \] Simplifying further: \[ = \frac{1}{100} \times 10^{-7} \times 2 \times 10 \times v \times 100 \times \frac{1}{8} \] \[ = 2.5 \times 10^{-7} = i \times R \] Solving for \( v \): \[ v = \frac{10 \times 10^{-6} \times 0.1}{2.5 \times 10^{-7}} = 4 \, \text{m/s} \]

Answer 2

Induced emf in AB = (\((\vec{V}\times \vec{B}).\vec{l}\)

\(B=\frac{l_0i}{2\pi r}=\frac{4\pi\times10^{-7}}{2\pi\times 4\times10^{-2}}=\frac{1}{2}\times10^{-4}T\)

emf in AB=e₁=\(B\times\frac{1}{2}\times2\times10^{-2}\times V\)

\(\Rightarrow \frac{V}{2}\times10^{-6}\,Volt\)

Induced emf in CD = \(e_2 = B\times \frac{1}{2}\times 2\times10^{-2}\times V\)

\(\Rightarrow \frac{\mu_0}{2\pi(8\times10^{-2})}\times\frac{1}{2}\times2\times10^{-2}\times V\)

\(\Rightarrow V\times\frac{1}{4}\times10^{-6}T\)

Emf in BC and AD are equal

emf in loop = \(e_1-e_2+e-e+e-e=e_1-e_2\)

\(V\times\frac{1}{2}\times10^{-6}-\frac{1}{4}\times10^{-6}\times V\)

\(=\frac{V}{4}\times10^{-6}\)

Resistance of loop = 0.1\(\Omega\)

Current in loop = I = \(\frac{V\times10^{-6}}{4\times0.1}=\frac{10}{4}\times V\mu A\)

\(\frac{10V}{4}=10\)

\(V=4\,ms\)