A rectangular frame ABCD made of a uniform metal wire has a...

${e}_{1} = A\dfrac{\text{d}B}{\text{d}t} = 1 \times \text{1} \times 1 = \text{1 V}$

${e}_{2} = 0 . 5V$

From loop $AEFDA$

$1\left({I}_{1} - {I}_{2}\right) - 1 + 3{I}_{1} = 0 ⇒ 4{I}_{1} - {I}_{2} = 1...(1)$

From loop $FCBEF$

$\text{0.5} - 2{I}_{2} + \left({I}_{1} - {I}_{2}\right) = 0 ⇒ - {I}_{1} + 3{I}_{2} = \text{0.5}...(2)$

From (1) and (2)

${I}_{1} = \dfrac{\text{7}}{\text{22}}\text{A,}{I}_{2} = \dfrac{3}{\text{11}}\text{A,}{I}_{1} - {I}_{2} = \dfrac{\text{1}}{\text{22}}\text{A}$

${I}_{\text{AE}} = \dfrac{\text{7}}{\text{22}}\text{A}$, ${I}_{\text{BE}} = \dfrac{3}{\text{11}}\text{A}$ , ${I}_{\text{EF}} = \dfrac{\text{1}}{\text{22}}\text{A}$

$\dfrac{{I}_{AE}}{{I}_{EF}} = 7$