Question

A conducting square frame of side '$a$' and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity '$V$'. The emf induced in the frame will be proportional to

• A. $\frac{1}{(2x+a)^2}$
• B. $\frac{1}{(2x-a)(2x+a)}$
• C. $\frac{1}{(2x-a)}$
• D. $\frac{1}{(2x-a)^2}$

Answer 1

Answer: (B)

Induced emf $\varepsilon=B_1 a v-B_2$ av
$=\frac{\mu_0 I}{2 \pi\left(x-\frac{a}{2}\right.}$ av $\left.-\frac{\mu_0 I}{2 \pi\left(x+\frac{1}{2}\right)}\right)$ av
$=\frac{\mu_0 I \text { av }}{2 \pi}\left(\frac{1}{x-\frac{1}{2}}-\frac{1}{x+\frac{4}{2}}\right)=$
$ \begin{array}{l} \frac{\mu_0 Iav }{2 \pi} \frac{ a }{(2 x-a)(2 x+a)} \\ \Rightarrow B \propto \frac{1}{(2 x-a)(2 x+a)} \end{array} $