Question

Figure shows a fixed square frame of wire having a total resistance r placed coplartarly with a long, straight wire. The wire carries a current i given by $i=i_{2}cos\left(2\pi t/T\right)$,Find (a) the flux of the magnetic field through the square frame, (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to 10 T.

Answer 1

As shown in the figure, ABCD is a square of side a. b is the distance from long wire carrying current $I$.

$i=i_0\cos \frac{2\pi t}{T}$

Let us consider a strip of thickness $dx$,distance x from wire.

Now magnetic field across $dx$ thickness will be,

$dB=\frac{{\mu }_{0}i}{2\pi x}$ into the plane.

Flux passing through thickness $dx$ is,

${\int }_0^{Q}dQ={\int }_b^{b+a}dB.\left(adx\right)[Q=B.ds]$

(a) $\varphi ={\int }_b^{b+a}\frac{{\mu }_{0}i}{2\pi x}dx=\frac{{\mu }_{0}i}{2\pi }\ln \frac{b+a}{b}$

(b) emf induced $e=-\frac{dQ}{dt}$

$e=-\frac{d}{dt}\left(\frac{{\mu }_0}{2\pi }\ln \frac{b+a}{b}{i}_0\cos \frac{2\pi t}{T}\right)$

$e=\frac{{\mu }_0{i}_0}{T}\ln \frac{b+a}{b}\sin \frac{2\pi t}{T}$

(c) Heat developed $dH=\frac{e^2}{r}dt$

$dH=\frac{{{\mu }_0}^2{i_0}^2}{T^{2}r}{\left(\ln \frac{b+a}{b}\right)}^2{\sin \left(\frac{2\pi t}{T}\right)}^{2}dt$

$=\frac{{{\mu }_0}^2{i_0}^2}{T^{2}r}{\left(\ln \frac{b+a}{b}\right)}^2\left[\frac{1-\cos \frac{4\pi t}{T}}{2}\right]dt$

${\int }_0^{H}dH={\int }_0^{10T}\frac{{{\mu }_0}^2{i_0}^2}{{2T}^{2}r}{\left(\ln \frac{b+a}{b}\right)}^2[1-\cos \frac{4\pi t}{T}]dt$

$H=K{[t-\frac{\sin \frac{4\pi t}{T}}{\frac{4\pi }{T}}]}_0^{10T}$

$H=K[10T-0-\frac{\sin 40\pi -0}{\frac{4\pi }{T}}]$ where K=$\frac{{\mu }^2{i_0}^2}{{2T}^{2}r}\ln {\left(\frac{b+a}{b}\right)}^2$

$H=K\times 10T$

$H$= $\frac{{5\mu }^2{i_0}^2}{Tr}\ln {\left(\frac{b+a}{b}\right)}^2$