As shown in the figure, a rectangular loop of a conducting wire is moving away with a constant velocity $v$ in a perpendiculat direction from a very long straight conductor carrying a steady current $l$. When the breadth of the rectangular loop is very small compared to its distance from the straight conductor, how does the emf. $E$ induced in the loop vary with time $t ?$

• Correct answer
  A

  $E \propto \frac{1}{t^{2}}$
• B

  $E \propto \frac{1}{t}$
• C

  $E \alpha-\ln (t)$
• D

  $E \propto \frac{1}{t^{3}}$

Solution

$\therefore$ Motional emf in the loop. $$ \begin{aligned} E &=E_{1}-E_{2}=vl l\left(B_{1}-B_{2}\right) \\ &=v\left(\frac{\mu_{0}}{2 \pi} \cdot \frac{I}{y}-\frac{\mu_{0} I}{2 \pi(y+b)}\right] \\ &=\frac{\mu_{0}}{2 \pi} \cdot I \cdot v l\left[\frac{y+b-y}{y(y+b)}\right]=\frac{\mu_{0}}{2 \pi} \cdot \frac{v I l}{y(y+b)} b \end{aligned} $$ $\because \quad b< < y$ $\therefore \quad E=\frac{\mu_{0}}{2 \pi} \cdot \frac{v l I}{y^{2}} b$ $\because v=$ constant $$ $$ Thus, $\quad E=\frac{\mu_{0}}{2 \pi} \cdot \frac{v l l}{v^{2} t^{2}} b$ Hence, $\quad E \propto \frac{1}{t^{2}}$

Hence, the correct answer is option: 'A'.