Two long currents 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 $$\theta$$ with the vertical. If wires have a mass $$\lambda$$ per unit length then the value of I is: ($$g$$ = gravitational acceleration)

Weight $$w = \lambda g$$ (Acts vertically downwards)

Magnetic Force $$F_m = \frac{\mu_0 I^2}{2 \pi d}$$

The total distance ($$d$$) between the two wires is $$d = 2L \sin \theta$$

Let $$T$$ be the tension per unit length of the thread. Resolving the tension into components:

Vertical equilibrium: $$T \cos \theta = \lambda g$$  --- (Eq. 1)

Horizontal equilibrium: $$T \sin \theta = F_m$$ --- (Eq. 2)

$$\tan \theta = \frac{F_m}{\lambda g}$$

$$\tan \theta = \frac{\mu_0 I^2}{2 \pi (2L \sin \theta) \lambda g}$$

$$\tan \theta = \frac{\mu_0 I^2}{2 \pi d \lambda g}$$

$$\frac{\sin \theta}{\cos \theta} = \frac{\mu_0 I^2}{4 \pi L \lambda g \sin \theta}$$

$$I^2 = \frac{4 \pi \lambda g L \sin^2 \theta}{\mu_0 \cos \theta}$$

$$I = \sqrt{\frac{4 \pi \lambda g L \sin^2 \theta}{\mu_0 \cos \theta}}$$

$$I = 2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}}$$