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 _______.
Answer (integer)
175
Solution
$\phi$ = $\overrightarrow B .\overrightarrow A$ = $$\left( {3\widehat i + 4\widehat k} \right).\left( {25\widehat i + 25\widehat k} \right)$$
<br><br>$\Rightarrow$ $\phi$ = (3 $\times$ 25) + (4 $\times$ 25) = 175 weber
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Faraday's Laws
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