JEE MAIN - Physics (2025 - 24th January Morning Shift - No. 25)
A square loop of sides $a=1 \mathrm{~m}$ is held normally in front of a point charge $\mathrm{q}=1 \mathrm{C}$ at a distance $\frac{\mathrm{a}}{2}$. The flux of the electric field through the shaded region is $\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_0} \frac{\mathrm{Nm}^2}{\mathrm{C}}$, where the value of p is ________ .
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Answer
48
Explanation
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Total flux through square $=\frac{\mathrm{q}}{\epsilon_0}\left(\frac{1}{6}\right)$
Lets divide square is 8 equal parts.
Flux is same for each part.
$\therefore$ Flux through shaded portion is $\frac{5}{8}$ (Total flux)
$$ =\frac{5}{8} \times \frac{\mathrm{q}}{\epsilon_0} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_0} $$
$\therefore$ Required Ans. is 48
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