Question

A conducting square frame of side $\text{a}$ and a long straight wire carrying current $\text{I}$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $\text{V}$ . The e.m.f induced in the frame (when the centre of the frame is at a distance $x$ from the wire) will be proportional to :

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

Answer 1

Answer: (D)

The potential difference across AB is
$V_A-V_B=B_{1}a.V.$
$\Rightarrow \frac{{\left(\mu \right)}_i}{2\pi \left(x-\frac{a}{2}\right)}aV$
The potential difference across CD is
$V_C-V_D=B_{2}a.V$
$B_2=\frac{{\left(\mu \right)}_{0}i}{2\pi \left(x+\frac{a}{2}\right)}$
$V_C-V_D=\frac{{\left(\mu \right)}_{0}i}{2\pi \left(x+\frac{a}{2}\right)}aV$
Net Potential difference $=\frac{{\left(\mu \right)}_{i}aV}{2\pi }\left(\frac{1}{x-\frac{a}{2}}-\frac{1}{x+\frac{a}{2}}\right)$
$(V_A-V_B)-(V_C-V_D)=\frac{\mu ia}{2\pi }(\frac{2a}{x^2-\frac{a^2}{4}})$
$\propto \frac{1}{4x^2-a^2}\propto \frac{1}{\left(2x+a\right)(2x-a)}$