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

Answer 1

The correct option is A

$\frac{1}{(2x-a)(2x+a)}$

See figure alongside.

Let x be the distance of the centre of the frame from the long straight wire carrying current I.

Consider the point P at a distance y from the long straight wire carrying current I.

Strength of magnetic induction at point P is given by

$B=\frac{{\mu }_0}{4\pi }\frac{21}{y}$

Integrating over y from y = $y=(x-\frac{a}{2})toy=(x+\frac{a}{2})$

We get

${\int }_{x-\frac{a}{2}}^{x+\frac{a}{2}}Bdy={\int }_{x-\frac{a}{2}}^{x+\frac{a}{2}}Bdy=\frac{{\mu }_0}{4\pi }\frac{2I}{y}dy=\frac{{\mu }_{0}Ia}{2\pi }ln[y{]}_{(x-\frac{a}{2})}^{(x+\frac{a}{2})}=\frac{{\mu }_{0}I}{2\pi }In\frac{x+\frac{a}{2}}{x-\frac{a}{2}}$

Total flux contained in the square frame is

$\varphi =\frac{{\mu }_{0}Ia}{2\pi }In\left[\frac{x+\frac{a}{2}}{x-\frac{a}{2}}\right]$

Rate of change of flux is

$\frac{d\varphi }{dt}=\frac{{\mu }_{0}Ia}{2\pi }\frac{d}{dt}\left[ln\left[\frac{x+\frac{a}{2}}{x-\frac{a}{2}}\right]\right]=\frac{{\mu }_{0}Ia}{2\pi }\left[\frac{x-\frac{a}{2}}{x+\frac{a}{2}}\right]\frac{d}{dt}\left[\frac{x+\frac{a}{2}}{x-\frac{a}{2}}\right]$
$=\frac{{\mu }_{0}Ia}{2\pi }\left[\frac{2x-a}{2x+a}\right]\frac{(x-\frac{a}{2})\frac{d}{dt}(x+\frac{a}{2})-(x+\frac{a}{2})\frac{d}{dt}(x-\frac{a}{2})}{(x-\frac{a}{2}{)}^2}$
$=\frac{{\mu }_{0}Ia}{2\pi }\left[\frac{2x-a}{2x+a}\right]\frac{4}{(x-\frac{a}{2}{)}^2}..[(x-\frac{a}{2})v-(x-\frac{a}{2})]$
$=\frac{2{\mu }_{0}Ia}{2\pi }\frac{1}{(2x-a)(2x+a)}v[-a]=-=\frac{2{\mu }_{0}I{a}^{2}v}{\pi }\frac{1}{(2x-a)(2x+a)}$
$\epsilon =-\frac{d\varphi }{dt}=\frac{2{\mu }_{0}I{a}^{2}v}{\pi }\frac{1}{(2x-a)(2x+a)}$
$\epsilon a\frac{1}{(2x-a)(2x+a)}$