Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle 'θ' with the vertical. If wires have mass λ per unit-length then the value of I is: ( g = gravitational acceleration) (a) 2√((π g L)/(μ₀)tan θ) (b) √((πλ g L)/(μ₀)tan θ) (c) sin θ√((πλ g L)/(μ₀cos θ)) (d) 2sin θ√((πλ g L)/(μ₀cos θ))

Answer: (d). Opposite currents repel. Separation r = 2Lsin θ; repulsion per unit length f = (μ₀ I²)/(2π (2Lsin θ)). Per unit length equilibrium: tan θ = (f)/(λ g) = (μ₀ I²)/(4π Lsin θ λ g) I² = (4πλ g Lsin θtan θ)/(μ₀) = (4πλ g Lsin ²θ)/(μ₀cos θ) I = 2sin θ√((πλ g L)/(μ₀cos θ)), so the answer is (d).

Two long current carrying thin wires, both with current I, are held… | JEE Main 2015

Q55JEE Main 2015Difficult

Force on Current-Carrying Conductors

Two long current carrying thin wires, both with current $I,$ are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $\text{'}\theta\text{'}$ with the vertical. If wires have mass $\lambda$ per unit-length then the value of $I$ is:

( $g =$ gravitational acceleration)

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Answer: (d)

Opposite currents repel. Separation $r = 2L\sin\theta$ ; repulsion per unit length $f = \frac{\mu_0 I^2}{2\pi (2L\sin\theta)}$ .
Per unit length equilibrium: $\tan\theta = \frac{f}{\lambda g} = \frac{\mu_0 I^2}{4\pi L\sin\theta\,\lambda g}$
$I^2 = \frac{4\pi\lambda g L\sin\theta\tan\theta}{\mu_0} = \frac{4\pi\lambda g L\sin^2\theta}{\mu_0\cos\theta}$
$I = 2\sin\theta\sqrt{\frac{\pi\lambda g L}{\mu_0\cos\theta}}$ , so the answer is (d).

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