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