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Two long conductors, separated by a distance 80 cm carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times -

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Question

Two long conductors, separated by a distance 80 cm carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 1.8 m. The new value of the force between them is ____________.

Options

  • - 8 F

  • F/3

  • - 8 F/9

  • - F/9

MCQ
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Solution

Two long conductors, separated by a distance 80 cm carry current I1 and I2 in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 1.8 m. The new value of the force between them is - 8 F/9.

Explanation:

Force between two long conductors carrying current,

`"F"= "mu"_0/(2pi) ("I"_1 "I"_2)/"d" l`   ....(i)

After carrying out changes,

`"F'" = mu_0/(2pi) ((-2"I"_1)( "I"_2))/("d'") l` 

From (i) and (ii),

`"F'"/"F" = (-2//"d'")/(1//"d") = -2("d"/"d'") = -2 (0.8/1.8) = -8/9`

`Rightarrow "F'" = -8/9  "F"`

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