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

Options

A. `B _(0) L ^(2) Wb.`
B. `2 B _(0) L ^(2) Wb.`
C. `sqrt2 B _(0) L ^(2) W b .`
D. `4 B _(0) L ^(2) Wb.`

Answer

B

Explanation

The loop can be considered in two planes:

(i) Plane of ABCDA is in X-Y plane. So its vector `vecA` is in Z-direction. Hence,
`A _(1) = | A| hatk = L ^(2) hatk`
(ii)Plane of DEFAD is in Y-Z plane
So `A _(2) = |A| hatk = L ^(2) hati`
`therefore A = A _(1) + A _(2) = L ^(2) (hati + hatk)`
`B = B _(0)( hati + hatk)`
So, `Q = B.A = B _(0) ( hati + hatk). L ^(2) (hati + hatk) = B _(0) L^(2)`
`[hati . hati+ hati.hatk+hatk.hati + hatk.hatk]`
`= B _(0) L ^(2) [1 + 0 + 0+ 1] (therefore cos 90^(@) = 0)`
`= 2 B _(0) L ^(2) Wb`