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 $1T{s}^{-1}$ , 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

Answer: 7

$e_1=A\frac{\text{d}B}{\text{d}t}=1\times \text{1}\times 1=\text{1 V}$
$e_2=0.5V$
From loop $AEFDA$
​ $1\left(I_1-I_2\right)-1+3I_1=0\Rightarrow 4I_1-I_2=\mathrm{1...}(1)$
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)
$I_1=\frac{\text{7}}{\text{22}}\text{A,}{I}_2=\frac{3}{\text{11}}\text{A,}{I}_1-I_2=\frac{\text{1}}{\text{22}}\text{A}$
$I_{\text{AE}}=\frac{\text{7}}{\text{22}}\text{A}$ , $I_{\text{BE}}=\frac{3}{\text{11}}\text{A}$ , $I_{\text{EF}}=\frac{\text{1}}{\text{22}}\text{A}$
$\frac{I_{AE}}{I_{EF}}=7$