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 ' $θ$ ' with the vertical. If wires have a mass $λ$ per unit length then the value of $I$ is:
( $g =$ gravitational acceleration)

Question

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 '

θ

' with the vertical. If wires have a mass

λ

per unit length then the value of

I

is:
(

g =

gravitational acceleration)

Aicharya · Accepted Answer

Two wires will repel each other due to magnetic force, then the magnetic force per unit length is,

\frac{d f}{d l} = \frac{μ_{0} I^{2}}{2 π (2 L sin θ)} = \frac{μ_{0} I^{2}}{4 π L sin θ}

. And mass per unit length of each wire

= \frac{d m}{d l} = λ

. So, the magnetic force on the total length

L

of the wire is

f_{m} = \frac{μ_{0} I^{2} L}{4 π L sin θ}

, and weight

= λ L g

. By equilibrium of wire,

T sin θ = f_{m} & T cos θ = W = λ l' g

\Rightarrow \frac{T sin θ}{T cos θ} = \frac{f_{m}}{mg} \Rightarrow f_{m} = λ l' g tan θ

\Rightarrow \frac{μ_{0} I^{2}}{4 π L sin θ} L = λ L g \frac{sin θ}{cos θ}

\Rightarrow I^{2} = \frac{λ g π L}{μ_{0} cos θ} 4 {sin}^{2} θ

⇒ I = 2 sin θ \sqrt{\frac{λ π gL}{μ_{0} cos θ}}

Two long currents carrying thin wires, both with current I , are held by insulating threads of length L and…