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QJEEM20Phc28d: JEE Main 2020 Physics - Class 12 JEEM20Phc28d Solution: A loop ABCDEFA of straight edges has six corner points A(0, 0,…

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 B→=(3i^+4k^)T\overrightarrow B = \left( {3\widehat i + 4\widehat k} \right)T . The quantity of flux through the loop ABCDEFA (in Wb) is _______.

Step-by-Step Verified Solution

ϕ\phi = B→.A→\overrightarrow B .\overrightarrow A = (3i^+4k^).(25i^+25k^)\left( {3\widehat i + 4\widehat k} \right).\left( {25\widehat i + 25\widehat k} \right)

⇒\Rightarrow ϕ\phi = (3 ×\times 25) + (4 ×\times 25) = 175 weber

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