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

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 $=\frac{\left(\mu \right)_{i} \, a V}{2 \, \pi } \, \left(\frac{1}{x \, - \, \frac{a}{2}} - \, \frac{1}{x \, + \, \frac{a}{2}}\right)$
$\left(\right. V_{A} - V_{B} \left.\right) - \left(\right. V_{C} - V_{D} \left.\right) = \frac{\mu i a}{2 \pi } \left(\right. \frac{2 a}{x^{2} - \frac{a^{2}}{4}} \left.\right)$
$ \propto \, \, \frac{1}{4 x^{2} - a^{2}} \, \, \, \propto \, \, \, \frac{1}{\left(2 x + a\right) \left(\right. 2 x - a \left.\right)} \, $