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A loop, made of straight edges has six c...

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:

A

`B_(0)L^(2)Wb`

B

`2B_(0)L^(2)Wb`

C

`sqrt(2)B_(0)L^(2)Wb`

D

`4B_(0)L^(2)Wb`

Text Solution

Verified by Experts

The correct Answer is:
B
`vecA=L^(2)hatk+L^(2)hati=L^(2)(hati+hatk)`
`therefore phi=vecB.vecA=B_(0)(hati+hatk).L^(2)(hati+hatk)=2B_(0)L^(2)Wb`
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