JEE Main2015PhysicsElectrostaticsActual
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)
Options
- Aπ λ g L μ 0 t a n θ
- Bs i n θ π λ g L μ 0   c o s θ
- C2 s i n θ π λ g L μ 0 c o s θ
- D2 π g L μ 0 t a n θ
Correct answer
C. 2 s i n θ π λ g L μ 0 c o s θ
Step-by-step solution
Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, d f d l = μ 0 I 2 2 π 2 L sin θ = μ 0 I 2 4 π L sin θ . And mass per unit length of each wire = d m d l = λ . So, the magnetic force on the total length L of the wire is f m = 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 ⇒ T