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Question:

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)

√πλgLμ₀tanθ

2sinθ√πλgLμ₀cosθ

sinθ√πλgLμ₀cosθ

2√πgLμ₀tanθ

Solution:

The wires repel each other with some force. The magnitude of this force can be given by
F = μ₀Il × B
where l is length of wire
B = μ₀I/2πr
where r is distance between two wires
r = 2Lsinθ
F = μ₀I²l/2π(2Lsinθ) .. (i)
Now the tension in the thread is keeping both wire from moving to sideways
From FBD of the wire
Tsinθ = F .. (ii)
Tcosθ = mg .. (iii)
dividing equation (ii) by (iii)
tanθ = F/mg
m = λL
I = 2sinθ√πλgLμ₀cosθ