A point charge \(q = 1\ \mu\text{C}\) is located at a distance \(2\text{ cm}\) from one end of a thin insulating wire of length \(10\text{ cm}\) having a charge \(Q = 24\ \mu\text{C}\), distributed uniformly along its length, as shown in figure. Force between \(q\) and wire is _______ N.
(Use \(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2\))

Explanation:
Let us calculate the electrostatic force between the point charge and the uniformly charged wire step-by-step using calculus:

1. Convert Given Values to SI Units:
- Point charge, \(q = 1\ \mu\text{C} = 10^{-6}\text{ C}\)
- Total charge on wire, \(Q = 24\ \mu\text{C} = 24 \times 10^{-6}\text{ C}\)
- Length of the wire, \(L = 10\text{ cm} = 10 \times 10^{-2}\text{ m} = 10^{-1}\text{ m}\)
- Distance of point charge from nearest end = \(2\text{ cm} = 2 \times 10^{-2}\text{ m}\)
- Distance from furthest end = \(2\text{ cm} + 10\text{ cm} = 12\text{ cm} = 12 \times 10^{-2}\text{ m}\)

2. Define Linear Charge Density (\(\lambda\)):
The charge per unit length of the wire is:
\(\lambda = \frac{Q}{L} = \frac{24 \times 10^{-6}}{10^{-1}}\text{ C/m}\)

3. Set Up the Integration:
Consider a small element of length \(dx\) on the wire at a distance \(x\) from the point charge \(q\). The small charge on this element is \(dq = \lambda dx\).
The small electrostatic force \(dF\) between \(q\) and \(dq\) is given by Coulomb's Law:
\(dF = \frac{k \cdot q \cdot dq}{x^2} = \frac{k \cdot q \cdot \lambda \cdot dx}{x^2}\)

Integrating this expression from the nearest end (\(x = 2\text{ cm}\)) to the farthest end (\(x = 12\text{ cm}\)):
\(F = \int_{2 \times 10^{-2}}^{12 \times 10^{-2}} \frac{k \cdot q \cdot \lambda}{x^2} dx\)
\(F = k \cdot q \cdot \lambda \left[ -\frac{1}{x} \right]_{2 \times 10^{-2}}^{12 \times 10^{-2}}\)
\(F = k \cdot q \cdot \lambda \left( \frac{1}{2 \times 10^{-2}} - \frac{1}{12 \times 10^{-2}} \right)\)
\(F = k \cdot q \cdot \lambda \cdot \left( \frac{5}{12 \times 10^{-2}} \right)\)

4. Substitute the Values:
\(F = (9 \times 10^9) \times (10^{-6}) \times \left( \frac{24 \times 10^{-6}}{10^{-1}} \right) \times \left( \frac{5}{12} \right) \times 10^2\)
\(F = 9 \times 24 \times \frac{5}{12} = 9 \times 2 \times 5 = 90\text{ N}\)

Hence, Option B is the correct choice.