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Current Electricity

Question
CBSEENPH12039487

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)

  • sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root
  • 2 space sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root
  • 2 square root of πgL over straight mu subscript 0 tanθ end root
  • square root of πλgL over straight mu subscript 0 tanθ end root

Solution

B.

2 space sin space straight theta space square root of fraction numerator πλgL over denominator straight mu subscript straight theta cos space straight theta end fraction end root

consider free body diagram of the wire
As the wires are in equilibrium. they must carry current in opposite direction.

Here, straight F subscript straight B space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator italic 2 pi d end fraction, where l is the length of each wire are d is a separation between wires.
 From the diagram, d = 2L sin θ
T = cos θ = mg = λlg 
(in vertical direction) ..... (i)
 straight T space sin space straight theta space equals space straight F subscript straight B space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator 4 straight pi space straight L space sin space straight theta end fraction
(In horizontal direction)  ..... (ii)
From eqs. (i) and (ii)
fraction numerator straight T space sin space straight theta over denominator straight T space cos space straight theta end fraction space equals space fraction numerator straight mu subscript 0 straight I squared l over denominator 4 πL space sin space straight theta space xλ l straight g end fraction
straight I space equals space square root of fraction numerator 4 πλLg space sin squared straight theta over denominator straight mu subscript straight theta space cos space straight theta end fraction end root
equals space 2 space sin space straight theta space square root of fraction numerator πλlg over denominator straight mu subscript straight o cos space straight theta end fraction end root