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A rectangular frame ABCD, made of a unif...

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.

Text Solution

Verified by Experts

The correct Answer is:
`I_(EA)=7/22 A; I_(BE)=3//11A; I_(FE)=1/22A`

`e_(1)=A(dB)/(dt)=1xx0.5xx1=1 V`
`e_(2)=0.5` volt
From loop `AEFD`
`1(I_(1)-I_(2))-1+3I_(1)=0`
`4I_(1)-I_(2)=1…(1)`
From loop `FCBEF`
`0.5-2I_(2)+1(I_(1)-I_(2))=0`
`-I_(1)+3I_(2)=0.5 …(2)`
From `(1)` and `(2)`
`I_(1)=7/22 Amp, I_(2)=3/11 Amp, I_(1)-I_(2)=4/11 Amp`
`I_(AE)=7/22Amp, I_(B)E=3/11 Amp, I_(EF)=4/11 Amp`
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