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 $\epsilon =B_{1}av-B_2$ av
$=\frac{{\mu }_{0}I}{2\pi (x-\frac{a}{2}}$ av $-\frac{{\mu }_{0}I}{2\pi \left(x+\frac{1}{2}\right)})$ av
$=\frac{{\mu }_{0}I\text{ av }}{2\pi }\left(\frac{1}{x-\frac{1}{2}}-\frac{1}{x+\frac{4}{2}}\right)=$
$\begin{array}{l}\frac{{\mu }_{0}Iav}{2\pi }\frac{a}{(2x-a)(2x+a)}\\ \Rightarrow B\propto \frac{1}{(2x-a)(2x+a)}\end{array}$