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

A, B

Explanation

The equivalent circuit is drawn in the adjecent figure

As the magnetic field increase in the downward direction, an induced emf will be produced in the AEFD as well as in the circuit EBCF such that the current flowing in the loop creates magnetic lines of force in the upward direction (to the plane of paper)
Thus, the current should flow in the anticlockwise direction in both the loops.
Induced emf in loop AEFD
`e=-(d phi)/(dt)=-(d)/(dt) BA =-A(dB)/(dt)=-1xx1=-volt`
Induced emf in loop EBCF
`e=-(d phi)/(dt)=-(d)/(dt)BA'=-A'(dB)/(dt)=-0.5xx1=-0.5 volt
Let the current flowing in the branch EADF be `(i_1)` and the current flowing in the branch FCBE be `(i_2)`. Applying junction law at F, we get current in branch FE to be `(i_(1)-i_(2))`
Applying Kirchhoff's law in loop EADFE
`-1xx(I_1)-1xx(I_1)+1-1xx(I_1)-1(i_(1)-i_(2))`=0`
`implies 4(i_1)-(i_2)=1`...(i)
Appying Kirchhoff's law in loop EBCDFE
`+0.5(i_1)-0.5+1(i_2)-1(i_(1)-i_(2))=0`
`-(i_1)+3(i_2)=0.5` ...(ii) Solving (i) and (ii)
`11(i_1)=3.5`
`implies (i_1)=3.5//11=7/22A`
Also `11(i_2)=3`
`implies (i_2)=3//11A=6/22A`
`:. Current in segment AE=(i_1)=7/22A`
`current in segment BE=(i_2)=6/22A`
Current in segment `EF=(i_(1)-i_(2))=1/22A`.