JEE Main 2026 — Physics Electromagnetism

Consider a small element of length $dx$dx on the wire at a distance $x$x from the point charge $q$q.

The linear charge density of the wire is $\lambda = \frac{Q}{L} = \frac{24 \times 10^{-6}}{0.1} = 240 \times 10^{-6}$λ=LQ​=0.124×10−6​=240×10−6 C/m.

The charge on the small element is $dQ = \lambda dx$dQ=λdx.

The electrostatic force between the point charge $q$q and the element $dQ$dQ is $dF = \frac{k q dQ}{x^2} = \frac{k q \lambda dx}{x^2}$dF=x2kqdQ​=x2kqλdx​.

The total force $F$F is obtained by integrating $dF$dF from $x = a$x=a to $x = a + L$x=a+L, where $a = 2$a=2 cm and $L = 10$L=10 cm.

$$F = \int_{0.02}^{0.12} \frac{k q \lambda}{x^2} dx = k q \lambda \left[ -\frac{1}{x} \right]_{0.02}^{0.12}$$F=∫0.020.12​x2kqλ​dx=kqλ[−x1​]0.020.12​

$$F = k q \lambda \left( \frac{1}{0.02} - \frac{1}{0.12} \right) = k q \frac{Q}{L} \left( \frac{0.12 - 0.02}{0.02 \times 0.12} \right)$$F=kqλ(0.021​−0.121​)=kqLQ​(0.02×0.120.12−0.02​)

$$F = \frac{k q Q}{L} \frac{L}{a(a+L)} = \frac{k q Q}{a(a+L)}$$F=LkqQ​a(a+L)L​=a(a+L)kqQ​

Substituting the given values: $k = 9 \times 10^9$k=9×109 N.m$^2$2/C$^2$2, $q = 1 \times 10^{-6}$q=1×10−6 C, $Q = 24 \times 10^{-6}$Q=24×10−6 C, $a = 0.02$a=0.02 m, and $a+L = 0.12$a+L=0.12 m.

$$F = \frac{9 \times 10^9 \times 1 \times 10^{-6} \times 24 \times 10^{-6}}{0.02 \times 0.12}$$F=0.02×0.129×109×1×10−6×24×10−6​

$$F = \frac{9 \times 24 \times 10^{-3}}{0.0024} = \frac{216 \times 10^{-3}}{2.4 \times 10^{-3}}$$F=0.00249×24×10−3​=2.4×10−3216×10−3​

$F = \frac{216}{2.4} = 90$F=2.4216​=90 N.