A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5,
0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is $\overrightarrow B = \left( {3\widehat i + 4\widehat k} \right)T$
. The quantity of
flux through the loop ABCDEFA (in Wb) is _______.
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Step-by-Step Solution
Step 1: Identify the vector form of the Magnetic Field
The magnetic field given is
\overrightarrow{B} = 3 \hat{i} + 4 \hat{k} \, \text{T} .
This is a constant vector field in the region.
Step 2: Determine the Plane (or Effective Area) of the Loop
The loop described by points
A(0, 0, 0) → B(5, 0, 0) → C(5, 5, 0) → D(0, 5, 0) → E(0, 5, 5) → F(0, 0, 5) → A(0, 0, 0)
can be split into two rectangular faces:
Rectangular face ABCD in the plane z = 0 with length 5 (AB) and breadth 5 (BC).
Rectangular face AEFD. However, observe that A(0,0,0) → D(0,5,0) → E(0,5,5) → F(0,0,5) forms a rectangle in a plane parallel to the y–z plane with length 5 (DE) and breadth 5 (EF). But since we move from D to E to F to A, that also makes a 5 × 5 rectangle.
To find the total area vector effectively enclosed by the loop, we note that each rectangle spans an area of 25. The net area vector from ABCD is
25 \hat{i} (pointing in the +x direction), and the net area vector from AEFD is
25 \hat{k} (pointing in the +z direction). Thus the total area vector of the loop can be written as
\overrightarrow{A} = 25 \hat{i} + 25 \hat{k} \, \text{m}^2.
Step 3: Compute the Dot Product for Magnetic Flux
Magnetic flux \phi through the loop is given by
\phi = \overrightarrow{B} \cdot \overrightarrow{A}.
Substituting the vectors, we get
\phi = \bigl(3 \hat{i} + 4 \hat{k}\bigr) \cdot \bigl(25 \hat{i} + 25 \hat{k}\bigr).
This dot product becomes
\phi = (3 \times 25) + (4 \times 25) = 75 + 100 = 175.
Step 4: State the Final Answer
The magnetic flux through the loop ABCDEFA is
175 \, \text{Wb}.
