A rectangular frame ABCD, made of a uniform metal wire, has a straight connection between E and F made of the samae wire, as shown in fig. AEFD is a square of side 1m, and EB=FC=0.5m. The entire circuit is placed in steadily increasing, uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is `1T//s`. The resistance per unit length of the wire is `1omega//m`. Find the magnitude and directions of the currents in the segments AE, BE and EF.

Answer

`(1)/(22)A (from F to E)`

Explanation

Loop `ABCD`: Area `= 1 xx 1 = 1 m^(2)` ltbr. Flux: `phi_(1) = B(Area) = B xx 1 = B`
Induced emf: `e_(1) = (dphi_(1))/(dt) = (dB)/(dt) = 1 V`

Loop `EBCF`: Area `= 1 xx ((1)/(2)) = (1)/(2)m^(2)`
Flux: `phi_(2) = B(Area) = (B)/(2)`
Induced emf: `e_(2) = (dphi)/(dt) = (1)/(2)(dB)/(dt) = (1)/(2) V`
Both emfs are anticlockwise sence as shown in figure.
Apply Kirchhoff's law:
`e_(1) = 1I + 1I + 1I - 1(I_(1) - I)` `rarr` `1 = 3I - I_(1) + I`
`rarr` `4I - I_(1) = 1` (i)
`e_(2) = 1(I_(1) - I) + (1)/(2)I_(1) + 1I + (1)/(2)I_(1)`
`rarr` `6I - 2I_(1) = 1` (ii)
Solve (i) and (ii) to get
`I = (7)/(22)A, I_(1) = (6)/(22)A`
So current in `EF`: `I_(1) - I = (6)/(22) - (7)/(22)A` (from `Eto F`)
`= (1)/(22)A` (from `F to E`)