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 gigure , With threads making an angle ` theta` with the vertical . If wires have mass ` lambda` per unit length then the value of `I` is :
A
`2 sqrt(pi g L)/( mu_(0)) tan theta`
B
` sqrt(pi lambda g L)/( mu_(0)) tan theta`
C
` sin theta sqrt(pi lambda g L)/( mu_(0) cos theta)`
D
`2 sin theta sqrt(pi lambda g L)/( mu_(0) cos theta)`
Text Solution
Verified by Experts
The correct Answer is:
D
Let us consider 'l' length of current carrying wire. At equilibrium `T cos theta = lambda gl` and ` t sin theta = (mu_(0))/( 2 pi) ( I xx Il)/( 2 L sin theta)` `[ :. (F_(B))/(l) = ( mu_(0))/( 4 pi) ( 2I xx I )/( 2 l sin theta)]` Therefore, ` I = 2 sin theta sqrt((pi lambda g L)/( mu_(0) cos theta))`
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