1.

A loop, made of straight edges has six corners at A(0,0,0), B(L,0,0) ,C(L,L,0), D(0,L,0), E(0,L,L) and F(0,0,L). A magnetic field vecB=B_0 (hati+hatk)T is present in the region. The flux passing through the loop ABCDEFA (in that order) is

Answer»

`B_0L^2Wb`
`2B_0L^2 Wb`
`sqrt2B_0L^2Wb`
`4B_0L^2Wb`

SOLUTION :Consider a plane is made by two loop.
Area of loop ABCDA in xy-plane , `vecA_1 = |A| hatk =L^2hatk`
Area of loop ADEFA in yz- plane
`vecA_2=|A|hati =L^2hati`
Total area of LOOPS are `vecA=vecA_1+vecA_2`
`THEREFORE vecA=L^2hatk+L^2hati`
Now flux linked with these loops is
`phi=vecA.vecB=(L^2hati+L^2hatk).B_0 (hati+hatk)` where `vecB=B_0 (hati+hatk)`
MAGNETIC flux ,
`phi=vecB.vecA`
`=B_0(hati+hatk).L^2(hati+hatk)`
`=B_0L^2(1+1)`
`=2B_0L^2`


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