A thin metallic wire having cross sectional area of 10⁻⁴ m² is used to make a ring of radius 30 cm. A positive charge of 2π C is uniformly distributed over the ring, while another positive charge of 30 pC is kept at the centre of the ring. The tension in the ring is ______ N; provided that the ring does not get deformed (neglect the influence of gravity). (given, (1)/(4πε₀) = 9 × 10⁹ SI units) Numerical answer type. Work it out, then open the answer.

Answer: 3. Small element R dθ with charge λ R dθ ( λ = Q/2π R) is pushed outward by the centre charge: (kqλ R dθ)/(R²). Balanced by tension: 2Tsin (dθ)/(2) = T dθ ⇒ T = (kqλ)/(R) = (kqQ)/(2π R²) T = 9 × 10⁹ × 30 × 10⁻¹² × 2π2π × 0.09 = (0.27)/(0.09) T = 3 N (the cross-sectional area is not needed).

A thin metallic wire having cross sectional area of 10⁻⁴ m² is used… | JEE Main 2024 (27 Jan, Shift 1)

Q36JEE Main 2024 · 27 JAN · SHIFT 1Difficult

Electric Charge & Coulomb's Law

A thin metallic wire having cross sectional area of $10^{-4}\text{ m}^{2}$ is used to make a ring of radius $30\text{ cm}$ . A positive charge of $2\pi\text{ C}$ is uniformly distributed over the ring, while another positive charge of 30 $\text{pC}$ is kept at the centre of the ring. The tension in the ring is ______ $\text{N}$ ; provided that the ring does not get deformed (neglect the influence of gravity). (given, $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ SI units)

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Small element $R\,d\theta$ with charge $\lambda R\,d\theta$ ( $\lambda = Q/2\pi R$ ) is pushed outward by the centre charge: $\dfrac{kq\lambda R\,d\theta}{R^2}$ .
Balanced by tension: $2T\sin\dfrac{d\theta}{2} = T\,d\theta \Rightarrow T = \dfrac{kq\lambda}{R} = \dfrac{kqQ}{2\pi R^2}$
$T = \dfrac{9 \times 10^9 \times 30 \times 10^{-12} \times 2\pi}{2\pi \times 0.09} = \dfrac{0.27}{0.09}$
$T = 3\text{ N}$ (the cross-sectional area is not needed).

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