Question

The induced emf in the loop, if the long wire carries a current of $50\, A$ and the loop has an instantaneous velocity $v=10\, m / s$ at the location $x=0.2\, m$ as shown in figure is $\varepsilon$ microvolt. Caluclate $3 \varepsilon$. (Take $a=0.1\, m$ )

Answer 1

Induced emf in the loop $=B_{1} v l-B_{2} v l$
$=\left(B_{1}-B_{2}\right) v l=\left[\frac{\mu_{0} I}{2 \pi x}-\frac{\mu_{0} I}{2 \pi(x+ a)}\right](v)(a)$
$=\frac{\mu_{0} I}{2 \pi}\left(\frac{a^{2} v}{x(x+ a)}\right)$
$=\frac{2 \times 10^{-7} \times 50 \times(0.1)^{2} \times 10}{0.2 \times 0.3}$
$=\frac{50}{3} \mu V$