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Magnetism And Matter

Question
CBSEENPH12039726

Two long conductors, separated by a distance d carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them increased to two times and its direction reversed. The distance is also increased to 3d. The new value of the force between them is

  • −2F 

  • F/3

  • −2F/3

  • -F/3

Solution

C.

−2F/3

Force acting between two current carrying conductors
straight F space equals space fraction numerator straight mu subscript 0 over denominator 2 straight pi end fraction fraction numerator straight I subscript 1 straight I subscript 2 over denominator straight d end fraction straight I space....... space left parenthesis straight i right parenthesis
where d = distance between the conductors,
          l = length of each conductor
straight F apostrophe space equals space fraction numerator straight mu subscript 0 over denominator 2 straight pi end fraction space fraction numerator left parenthesis negative 2 straight I subscript 1 right parenthesis left parenthesis straight I subscript 2 right parenthesis over denominator left parenthesis 3 straight d right parenthesis end fraction. space straight I
space equals space minus space fraction numerator straight mu subscript 0 over denominator 2 straight pi end fraction space fraction numerator left parenthesis negative 2 straight I subscript 1 right parenthesis left parenthesis straight I subscript 2 right parenthesis over denominator left parenthesis 3 straight d right parenthesis end fraction. straight I
space equals space minus space fraction numerator straight mu subscript 0 over denominator 2 straight pi end fraction fraction numerator 2 straight I subscript 1 straight I subscript 2 over denominator 3 straight d end fraction. space straight I space..... space left parenthesis ii right parenthesis
space thus comma space from space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space
fraction numerator straight F apostrophe over denominator straight F end fraction space equals space minus space 2 over 3
straight F apostrophe space equals space minus 2 straight F divided by 3