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Question

(D) The loop lies in the $xy$-plane with corners at $(L, L, 0)$,$(-L, L, 0)$,$(-L, -L, 0)$,and $(L, -L, 0)$.The side length of the square loop is $2L$

Vedclass · Accepted Answer

$4 B_0L^2 	ext{ Wb}$

Vedclass · Answer

(D) The loop lies in the $xy$-plane with corners at $(L, L, 0)$,$(-L, L, 0)$,$(-L, -L, 0)$,and $(L, -L, 0)$.The side length of the square loop is $2L$.Therefore,the area of the loop is $A = (2L) 	imes (2L) = 4L^2$.The area vector $\vec A$ is perpendicular to the $xy$-plane,so $\vec A = 4L^2 \hat k$.The magnetic field is given by $\vec B = B_0(\hat i + \hat k)$.The magnetic flux $\phi$ is given by the dot product of the magnetic field and the area vector:$\phi = \vec B \cdot \vec A$$\phi = B_0(\hat i + \hat k) \cdot (4L^2 \hat k)$$\phi = 4B_0L^2 (\hat i \cdot \hat k + \hat k \cdot \hat k)$Since $\hat i \cdot \hat k = 0$ and $\hat k \cdot \hat k = 1$,we get:$\phi = 4B_0L^2 (0 + 1) = 4B_0L^2 	ext{ Wb}$.

$A$ loop,made of straight edges,has four corners at $A(L, L, 0)$,$B(-L, L, 0)$,$C(-L, -L, 0)$,and $D(L, -L, 0)$. $A$ magnetic field $\vec B = B_0(\hat i + \hat k) \text{ T}$ is present in the region. The magnetic flux passing through the loop $ABCD$ is:

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