Question

A rectangular frame $ABCD$ made of a uniform metal wire has a straight connection between $E$ & $F$ made of the same wire as shown in the figure. $AEFD$ is a square of side $1m$ & $EB=FC=0.5m$ . The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper. The rate of change of the magnetic field is $1Ts^{- 1}$ , the resistance per unit length of the wire is $1\Omega m^{- 1}$ . If the currents in the segments $AE$ and $EF$ are $I_{AE}$ and $I_{EF}$ respectively, then what is the value of $\frac{I_{AE}}{I_{EF}}$ ?

Answer 1

$e_{1}=A\frac{\text{d} B}{\text{d} t}=1\times \text{1}\times 1=\text{1 V}$
$e_{2}=0.5V$
From loop $AEFDA$
​ $1\left(I_{1} - I_{2}\right)-1+3I_{1}=0\Rightarrow 4I_{1}-I_{2}=1...(1)$
From loop $FCBEF$
$\text{0.5}-2I_{2}+\left(I_{1} - I_{2}\right)=0\Rightarrow -I_{1}+3I_{2}=\text{0.5}...(2)$
From (1) and (2)
$I_{1}=\frac{\text{7}}{\text{22}}\text{A,}I_{2}=\frac{3}{\text{11}}\text{A,}I_{1}-I_{2}=\frac{\text{1}}{\text{22}}\text{A}$
$I_{\text{AE}}=\frac{\text{7}}{\text{22}}\text{A}$ , $I_{\text{BE}}=\frac{3}{\text{11}}\text{A}$ , $I_{\text{EF}}=\frac{\text{1}}{\text{22}}\text{A}$
$\frac{I_{AE}}{I_{EF}}=7$