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Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density $λ$ passes through its center and is perpendicular to the plane of wire

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$\frac{λq}{4 π^{2} ε_{0} R}$

$\frac{λq}{4 {πε}_{0} R}$

$\frac{λq}{2 π^{2} ε_{0} R}$

$\frac{λq}{π^{2} ε_{0} R}$

answer is B.

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Detailed Solution

Due to symmetry $F_{n e t} = \int d q E c o s θ$

$= \int_{- π / 2}^{π / 2} (\frac{q}{π R}) R d θ \frac{λ}{2 π ε \cdot R} c o s θ$

$= \frac{λ q}{2 π^{2} ε_{0} R} \int_{- π / 2}^{π / 2} c o s θ d θ$

$= \frac{λq}{2 π^{2} ε_{0} R} [sinθ]_{- π / 2}^{π / 2} = \frac{λq}{2 π^{2} ε_{0} R} [1 - (- 1)] = \frac{λq}{π^{2} ε_{0} R}$

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