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(A) The magnetic field $B$ at a distance $y$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_{0} I}{2 \pi y}$.Let the loop h

Vedclass · Accepted Answer

$E \propto \frac{1}{t^{2}}$

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(A) The magnetic field $B$ at a distance $y$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_{0} I}{2 \pi y}$.Let the loop have length $l$ and breadth $b$. The loop moves with velocity $v$ away from the wire. Let $y$ be the distance of the inner edge of the loop from the wire at $t=0$. At time $t$, the distance is $y(t) = y_0 + vt$.Since the breadth $b$ is very small $(b \ll y)$, the magnetic field across the loop can be approximated as uniform, and the induced emf $E$ is given by the difference in motional emf across the two sides of length $l$:$E = E_1 - E_2 = v l (B_1 - B_2) = v l \left( \frac{\mu_{0} I}{2 \pi y} - \frac{\mu_{0} I}{2 \pi (y+b)} \right)$$E = \frac{\mu_{0} I v l}{2 \pi} \left( \frac{y+b-y}{y(y+b)} \right) = \frac{\mu_{0} I v l b}{2 \pi y(y+b)}$Since $b \ll y$, we have $y+b \approx y$, so $E \approx \frac{\mu_{0} I v l b}{2 \pi y^2}$.Since $y = y_0 + vt$, for large $t$, $y \approx vt$. Thus, $E \propto \frac{1}{(vt)^2} \propto \frac{1}{t^2}$.

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