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

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 = \(\frac{V\times10^{-6}}{4\times0.1}=\frac{10}{4}\times V\mu A\)
\(\frac{10V}{4}=10\)
\(V=4\,ms\)

Answer 2

The two sides perpendicular to the wire don't contribute to the net electromotive force (emf). For the parallel sides:

\(\vec{E} = \vec{B} \times \vec{V}\) 
=\(\frac{\mu_0 l}{2\pi} \times \text{cross section of } V\)

That is, net \(emf =\)\((E_1 \cos 60^\circ - E_2 \cos 60^\circ) \times \text{width}\)

=\(\frac{1}{2} \times \frac{2}{100} \times \frac{\mu_0 l v}{2\pi} \times \left(\frac{1}{{\frac{4}{100}}} - \frac{1}{{\frac{8}{100}}}\right)\)

=\(2.5 \times 10^{-7} i_R\)

\(v = \frac{10 \times 10^{-6} \times 0.12}{2.5 \times 10^{-7}} = 4 \, \text{m/s}\)