Verified Solution • 2025–26 Session moderate

QJEEM24Ph6d44: JEE Main 2024 Physics - Class 12 JEEM24Ph6d44 Solution: An electric field is given by . The electric flux through a…

An electric field is given by (6i^+5j^+3k^)N/C(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C} . The electric flux through a surface area 30i^ m230 \hat{i} \mathrm{~m}^2 lying in YZ-plane (in SI unit) is :

Multiple Choice Options

  • A 60
  • B 90
  • C 180
  • D 150

Step-by-Step Verified Solution

amp;E→=6i^+5j^+3k^amp;A→=30i^amp;ϕ=E→⋅A→amp;ϕ=(6i^+5j^+3k^)⋅(30i^)amp;ϕ=6×30=180\begin{aligned}& \overrightarrow{\mathrm{E}}=6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}} \\& \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\& \phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}}) \\& \phi=6 \times 30=180\end{aligned}

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